牛客网 数据库实战60题SQL
二刷牛客网 数据库实战60题
第一题:
题目描述
查找最晚入职员工的所有信息
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`))
输入描述:
[code]无
输出描述:
[code]
emp_no | birth_date | first_name | last_name | gender | hire_date |
---|---|---|---|---|---|
10008 | 1958-02-19 | Saniya | Kalloufi | M | 1994-09-15 |
代码:
[code]SELECT * FROM employees ORDER BY hire_date DESC limit 0,1
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第二题:
题目描述
查找入职员工时间排名倒数第三的员工所有信息
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
输入描述:
[code]无
输出描述:
[code]
emp_no | birth_date | first_name | last_name | gender | hire_date |
---|---|---|---|---|---|
10005 | 1955-01-21 | Kyoichi | Maliniak | M | 1989-09-12 |
代码:
[code]SELECT * FROM employees ORDER BY hire_date DESC LIMIT 2,1
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第三题:
题目描述
查找各个部门当前(to_date='9999-01-01')领导当前薪水详情以及其对应部门编号dept_no
CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输入描述:
[code]无
输出描述:
[code]
emp_no | salary | from_date | to_date | dept_no |
---|---|---|---|---|
10002 | 72527 | 2001-08-02 | 9999-01-01 | d001 |
10004 | 74057 | 2001-11-27 | 9999-01-01 | d004 |
10005 | 94692 | 2001-09-09 | 9999-01-01 | d003 |
10006 | 43311 | 2001-08-02 | 9999-01-01 | d002 |
10010 | 94409 | 2001-11-23 | 9999-01-01 | d006 |
代码:
[code]SELECT s.*,dm.dept_no FROM salaries AS s, dept_manager AS dm WHERE s.to_date= '9999-01-01' AND dm.to_date= '9999-01-01' AND s.emp_no=dm.emp_no
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第四题:
题目描述
查找所有已经分配部门的员工的last_name和first_name
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
输入描述:
[code]无
输出描述:
[code]
last_name | first_name | dept_no |
---|---|---|
Facello | Georgi | d001 |
省略 | 省略 | 省略 |
Piveteau | Duangkaew | d006 |
代码:
[code]SELECT e.last_name,e.first_name,de.dept_no FROM dept_emp de,employees e WHERE de.emp_no=e.emp_no
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第五题:
题目描述
查找所有员工的last_name和first_name以及对应部门编号dept_no,也包括展示没有分配具体部门的员工
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
输入描述:
[code]无
输出描述:
[code]
last_name | first_name | dept_no |
---|---|---|
Facello | Georgi | d001 |
省略 | 省略 | 省略 |
Sluis | Mary | NULL(在sqlite中此处为空,MySQL为NULL) |
代码:
[code]SELECT e.last_name,e.first_name,de.dept_no FROM employees AS e LEFT JOIN dept_emp AS de ON e.emp_no=de.emp_no
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第六题:
题目描述
查找所有员工入职时候的薪水情况,给出emp_no以及salary, 并按照emp_no进行逆序
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输入描述:
[code]无
输出描述:
[code]
emp_no | salary |
---|---|
10011 | 25828 |
省略 | 省略 |
10001 | 60117 |
代码:
[code]SELECT e.emp_no,s.salary FROM employees AS e, salaries AS s WHERE e.emp_no=s.emp_no AND e.hire_date=s.from_date ORDER BY e.emp_no DESC
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第七题:
题目描述
查找薪水涨幅超过15次的员工号emp_no以及其对应的涨幅次数t
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输入描述:
[code]无
输出描述:
[code]
emp_no | t |
---|---|
10001 | 17 |
10004 | 16 |
10009 |
18 |
代码:
[code]SELECT s.emp_no,COUNT(emp_no) AS t FROM salaries AS s GROUP BY emp_no HAVING t>15
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第八题:
题目描述
找出所有员工当前(to_date='9999-01-01')具体的薪水salary情况,对于相同的薪水只显示一次,并按照逆序显示
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输入描述:
[code]无
输出描述:
[code]
salary |
---|
94692 |
94409 |
88958 |
88070 |
74057 |
72527 |
59755 |
43311 |
25828 |
代码:
[code]SELECT DISTINCT salary FROM salaries WHERE to_date='9999-01-01' ORDER BY salary DESC
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第九题
题目描述
获取所有部门当前manager的当前薪水情况,给出dept_no, emp_no以及salary,当前表示to_date='9999-01-01'
CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输入描述:
[code]无
输出描述:
[code]
dept_no | emp_no | salary |
---|---|---|
d001 | 10002 | 72527 |
d004 | 10004 | 74057 |
d003 | 10005 | 94692 |
d002 | 10006 | 43311 |
d006 | 10010 |
94409 |
代码:
[code]SELECT dm.dept_no,dm.emp_no,s.salary FROM dept_manager AS dm, salaries AS s WHERE dm.to_date='9999-01-01' AND s.to_date='9999-01-01' AND dm.emp_no=s.emp_no
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第十题:
题目描述
获取所有非manager的员工emp_no
CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
输入描述:
[code]无
输出描述:
[code]
emp_no |
---|
10001 |
10003 |
10007 |
10008 |
10009 |
10011 |
代码:
[code]SELECT e.emp_no FROM employees AS e WHERE e.emp_no NOT IN(SELECT dm.emp_no FROM dept_manager AS dm)
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第十一题:
题目描述
获取所有员工当前的manager,如果当前的manager是自己的话结果不显示,当前表示to_date='9999-01-01'。
结果第一列给出当前员工的emp_no,第二列给出其manager对应的manager_no。
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
输入描述:
[code]无
输出描述:
[code]
emp_no | manager_no |
---|---|
10001 | 10002 |
10003 | 10004 |
10009 | 10010 |
代码:
[code]SELECT de.emp_no,dm.emp_no AS manager_no FROM dept_emp AS de,dept_manager AS dm WHERE de.to_date='9999-01-01' AND dm.to_date='9999-01-01' AND dm.emp_no<>de.emp_no AND de.dept_no=dm.dept_no
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第十二题:
题目描述
获取所有部门中当前员工薪水最高的相关信息,给出dept_no, emp_no以及其对应的salary
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输入描述:
[code]无
输出描述:
[code]
dept_no | emp_no | salary |
---|---|---|
d001 | 10001 | 88958 |
d002 | 10006 | 43311 |
d003 | 10005 | 94692 |
d004 | 10004 | 74057 |
d005 | 10007 | 88070 |
d006 | 10009 |
95409 |
代码:
[code]SELECT de.dept_no,de.emp_no,MAX(s.salary) AS salary FROM dept_emp AS de, salaries AS s WHERE de.emp_no=s.emp_no AND de.to_date='9999-01-01' AND s.to_date='9999-01-01' GROUP BY de.dept_no
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第十三题:
题目描述
从titles表获取按照title进行分组,每组个数大于等于2,给出title以及对应的数目t。
CREATE TABLE IF NOT EXISTS "titles" (
`emp_no` int(11) NOT NULL,
`title` varchar(50) NOT NULL,
`from_date` date NOT NULL,
`to_date` date DEFAULT NULL);
输入描述:
[code]无
输出描述:
[code]
title | t |
---|---|
Assistant Engineer | 2 |
Engineer | 4 |
省略 | 省略 |
Staff | 3 |
代码:
[code]SELECT title, COUNT(title) AS t FROM titles GROUP BY title HAVING t>=2
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第十四题:
题目描述
从titles表获取按照title进行分组,每组个数大于等于2,给出title以及对应的数目t。
注意对于重复的emp_no进行忽略。
CREATE TABLE IF NOT EXISTS `titles` (
`emp_no` int(11) NOT NULL,
`title` varchar(50) NOT NULL,
`from_date` date NOT NULL,
`to_date` date DEFAULT NULL);
输入描述:
[code]无
输出描述:
[code]
title | t |
---|---|
Assistant Engineer | 2 |
Engineer | 3 |
省略 | 省略 |
Staff |
3 |
代码:
[code]SELECT title,COUNT(DISTINCT emp_no) AS T FROM titles GROUP BY title HAVING t>=2
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第十五题:
题目描述
查找employees表所有emp_no为奇数,且last_name不为Mary的员工信息,并按照hire_date逆序排列
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
输入描述:
[code]无
输出描述:
[code]
emp_no | birth_date | first_name | last_name | gender | hire_date |
---|---|---|---|---|---|
10011 | 1953-11-07 | Mary | Sluis | F | 1990-01-22 |
10005 | 1955-01-21 | Kyoichi | Maliniak | M | 1989-09-12 |
10007 | 1957-05-23 | Tzvetan | Zielinski | F | 1989-02-10 |
10003 | 1959-12-03 | Parto | Bamford | M | 1986-08-28 |
10001 | 1953-09-02 | Georgi | Facello | M | 1986-06-26 |
10009 | 1952-04-19 | Sumant | Peac | F | 1985-02-18 |
代码:
[code]SELECT * FROM employees WHERE emp_no%2=1 AND last_name<>'Mary' ORDER BY hire_date DESC
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第十六题:
题目描述
统计出当前各个title类型对应的员工当前(to_date='9999-01-01')薪水对应的平均工资。结果给出title以及平均工资avg。
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
CREATE TABLE IF NOT EXISTS "titles" (
`emp_no` int(11) NOT NULL,
`title` varchar(50) NOT NULL,
`from_date` date NOT NULL,
`to_date` date DEFAULT NULL);
输入描述:
[code]无
输出描述:
[code]
title | avg |
---|---|
Engineer | 94409.0 |
Senior Engineer | 69009.2 |
Senior Staff | 91381.0 |
Staff | 72527.0 |
代码:
[code]SELECT t.title,AVG(s.salary) FROM salaries AS s,titles AS t WHERE s.emp_no=t.emp_no AND s.to_date='9999-01-01' AND t.to_date='9999-01-01' GROUP BY t.title
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第十七题:
题目描述
获取当前(to_date='9999-01-01')薪水第二多的员工的emp_no以及其对应的薪水salary
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输入描述:
[code]无
输出描述:
[code]
emp_no | salary |
---|---|
10009 | 94409 |
代码:
[code]SELECT emp_no,salary FROM salaries WHERE to_date='9999-01-01' ORDER BY salary DESC LIMIT 1,1
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第十八题:
题目描述
查找当前薪水(to_date='9999-01-01')排名第二多的员工编号emp_no、薪水salary、last_name以及first_name,不准使用order by
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输入描述:
[code]无
输出描述:
[code]
emp_no | salary | last_name | first_name |
---|---|---|---|
10009 | 94409 | Peac | Sumant |
代码:
[code]SELECT e.emp_no,s.salary,e.last_name,e.first_name FROM employees AS e , salaries AS s WHERE e.emp_no=s.emp_no AND s.to_date='9999-01-01' AND s.salary=(SELECT MAX(salary) FROM salaries WHERE to_date='9999-01-01' AND salary<>(SELECT MAX(salary) FROM salaries WHERE to_date='9999-01-01' ) )
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第十九题:
题目描述
查找所有员工的last_name和first_name以及对应的dept_name,也包括暂时没有分配部门的员工
CREATE TABLE `departments` (
`dept_no` char(4) NOT NULL,
`dept_name` varchar(40) NOT NULL,
PRIMARY KEY (`dept_no`));
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
输入描述:
[code]无
输出描述:
[code]
last_name | first_name | dept_name |
---|---|---|
Facello | Georgi | Marketing |
省略 | 省略 | 省略 |
Sluis | Mary | NULL |
代码:
[code]SELECT e.last_name,e.first_name,d.dept_name FROM employees AS e LEFT JOIN dept_emp AS de ON e.emp_no=de.emp_no LEFT JOIN departments AS d ON d.dept_no=de.dept_no
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第二十题:
题目描述
查找员工编号emp_no为10001其自入职以来的薪水salary涨幅值growth
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输入描述:
[code]无
输出描述:
[code]
growth |
---|
28841 |
代码:
[code]SELECT MAX(salary)-MIN(salary) AS growth FROM salaries WHERE emp_no='10001'
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第二十一题:
题目描述
查找所有员工自入职以来的薪水涨幅情况,给出员工编号emp_no以及其对应的薪水涨幅growth,并按照growth进行升序
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输入描述:
[code]无
输出描述:
[code]
emp_no | growth |
---|---|
10011 | 0 |
省略 | 省略 |
10010 | 54496 |
10004 |
34003 |
代码:
[code]SELECT e.emp_no,(b.salary - c.salary) AS growth FROM employees AS e INNER JOIN salaries AS b ON e.emp_no=b.emp_no AND b.to_date='9999-01-01' INNER JOIN salaries AS c ON e.emp_no=c.emp_no AND e.hire_date=c.from_date ORDER BY growth ASC;
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第二十二题:
题目描述
统计各个部门对应员工涨幅的次数总和,给出部门编码dept_no、部门名称dept_name以及次数sum
CREATE TABLE `departments` (
`dept_no` char(4) NOT NULL,
`dept_name` varchar(40) NOT NULL,
PRIMARY KEY (`dept_no`));
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输入描述:
[code]无
输出描述:
[code]
dept_no | dept_name | sum |
---|---|---|
d001 | Marketing | 24 |
d002 | Finance | 14 |
d003 | Human Resources | 13 |
d004 | Production | 24 |
d005 | Development | 25 |
d006 | Quality Management | 25 |
代码:
[code]SELECT d.dept_no,d.dept_name,COUNT(s.salary) AS sum FROM dept_emp AS de JOIN salaries AS s ON de.emp_no=s.emp_no JOIN departments AS d ON d.dept_no=de.dept_no GROUP BY d.dept_no,d.dept_name;
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第二十三题:
题目描述
对所有员工的当前(to_date='9999-01-01')薪水按照salary进行按照1-N的排名,相同salary并列且按照emp_no升序排列
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输入描述:
[code]无
输出描述:
[code]
emp_no | salary | rank |
---|---|---|
10005 | 94692 | 1 |
10009 | 94409 | 2 |
10010 | 94409 | 2 |
10001 | 88958 | 3 |
10007 | 88070 | 4 |
10004 | 74057 | 5 |
10002 | 72527 | 6 |
10003 | 43311 | 7 |
10006 | 43311 | 7 |
10011 | 25828 | 8 |
代码:
[code]SELECT A.emp_no,B.salary, COUNT(DISTINCT B.salary) AS rank FROM salaries AS A , salaries AS B WHERE A.to_date='9999-01-01' AND B.to_date='9999-01-01' AND A.salary<=B.salary GROUP BY A.emp_no,A.salary ORDER BY A.salary DESC,A.emp_no ASC
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第二十四题:
题目描述
获取所有非manager员工当前的薪水情况,给出dept_no、emp_no以及salary ,当前表示to_date='9999-01-01'
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输入描述:
[code]无
输出描述:
[code]
dept_no | emp_no | salary |
---|---|---|
d001 | 10001 | 88958 |
d004 | 10003 | 43311 |
d005 | 10007 | 88070 |
d006 | 10009 | 95409 |
代码:
[code]SELECT de.dept_no,de.emp_no,s.salary FROM employees AS e JOIN dept_emp AS de ON e.emp_no=de.emp_no JOIN salaries AS s ON e.emp_no=s.emp_no WHERE de.to_date='9999-01-01' AND s.to_date='9999-01-01' AND de.emp_no NOT IN(SELECT m.emp_no FROM dept_manager m WHERE de.dept_no = m.dept_no AND m.to_date = '9999-01-01')
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第二十五题:
题目描述
获取员工其当前的薪水比其manager当前薪水还高的相关信息,当前表示to_date='9999-01-01',
结果第一列给出员工的emp_no,
第二列给出其manager的manager_no,
第三列给出该员工当前的薪水emp_salary,
第四列给该员工对应的manager当前的薪水manager_salary
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输入描述:
[code]无
输出描述:
[code]
emp_no | manager_no | emp_salary | manager_salary |
---|---|---|---|
10001 | 10002 | 88958 | 72527 |
10009 | 10010 | 95409 | 94409 |
代码:
[code]SELECT de.emp_no,dm.emp_no,se.salary,sm.salary FROM dept_emp AS de JOIN dept_manager AS dm ON de.dept_no=dm.dept_no INNER JOIN salaries AS se ON se.emp_no=de.emp_no INNER JOIN salaries AS sm ON sm.emp_no=dm.emp_no WHERE de.to_date='9999-01-01' AND se.to_date='9999-01-01' AND dm.to_date='9999-01-01' AND sm.to_date='9999-01-01' AND se.salary>sm.salary
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第二十六题:
题目描述
汇总各个部门当前员工的title类型的分配数目,结果给出部门编号dept_no、dept_name、其当前员工所有的title以及该类型title对应的数目count
CREATE TABLE `departments` (
`dept_no` char(4) NOT NULL,
`dept_name` varchar(40) NOT NULL,
PRIMARY KEY (`dept_no`));
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE IF NOT EXISTS `titles` (
`emp_no` int(11) NOT NULL,
`title` varchar(50) NOT NULL,
`from_date` date NOT NULL,
`to_date` date DEFAULT NULL);
输入描述:
[code]无
输出描述:
[code]
dept_no | dept_name | title | count |
---|---|---|---|
d001 | Marketing | Senior Engineer | 1 |
d001 | Marketing | Staff | 1 |
d002 | Finance | Senior Engineer | 1 |
d003 | Human Resources | Senior Staff | 1 |
d004 | Production | Senior Engineer | 2 |
d005 | Development | Senior Staff | 1 |
d006 | Quality Management | Engineer | 2 |
d006 | Quality Management | Senior Engineer | 1 |
代码:
[code]SELECT d.dept_no,d.dept_name,t.title,COUNT(t.title) FROM departments AS d,dept_emp AS de,titles AS t WHERE de.dept_no=d.dept_no AND t.emp_no=de.emp_no AND de.to_date='9999-01-01' AND t.to_date='9999-01-01' GROUP BY d.dept_no,d.dept_name,t.title
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第二十七题:
题目描述
给出每个员工每年薪水涨幅超过5000的员工编号emp_no、薪水变更开始日期from_date以及薪水涨幅值salary_growth,并按照salary_growth逆序排列。
提示:在sqlite中获取datetime时间对应的年份函数为strftime('%Y', to_date)
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输入描述:
[code]无
输出描述:
[code]
emp_no | from_date | salary_growth |
---|---|---|
10003 | 1995-12-03 | 24178 |
10008 | 1998-03-11 | 20843 |
10008 | 2000-03-10 | 5997 |
代码:
[code]SELECT s2.emp_no, s2.from_date, (s2.salary - s1.salary) AS salary_growth FROM salaries AS s1 INNER JOIN salaries AS s2 ON s1.emp_no = s2.emp_no WHERE salary_growth > 5000 AND (strftime("%Y", s2.from_date) - strftime("%Y", s1.from_date) =1 OR strftime("%Y", s2.to_date) - strftime("%Y", s1.to_date) = 1) ORDER BY salary_growth DESC
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第二十八题:
题目描述
film表
字段 | 说明 |
film_id | 电影id |
title | 电影名称 |
description | 电影描述信息 |
CREATE TABLE IF NOT EXISTS film (
film_id smallint(5) NOT NULL DEFAULT '0',
title varchar(255) NOT NULL,
description text,
PRIMARY KEY (film_id));
category表
字段 | 说明 |
category_id | 电影分类id |
name | 电影分类名称 |
last_update | 电影分类最后更新时间 |
CREATE TABLE category (
category_id tinyint(3) NOT NULL ,
name varchar(25) NOT NULL, `last_update` timestamp,
PRIMARY KEY ( category_id ));
film_category表
字段 | 说明 |
film_id | 电影id |
category_id | 电影分类id |
last_update | 电影id和分类id对应关系的最后更新时间 |
CREATE TABLE film_category (
film_id smallint(5) NOT NULL,
category_id tinyint(3) NOT NULL, `last_update` timestamp);
查找描述信息中包括robot的电影对应的分类名称以及电影数目,而且还需要该分类对应电影数量>=5部
输入描述:
[code]无
输出描述:
[code]无
代码:
[code]SELECT c.name, COUNT(f.film_id) FROM film AS f,category AS c,film_category AS fc, (SELECT category_id FROM film_category GROUP BY category_id HAVING COUNT(film_id)>=5 ) AS new_fc WHERE f.description LIKE '%robot%' AND fc.category_id=c.category_id AND fc.film_id=f.film_id AND c.category_id=new_fc.category_id
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第二十九题:
题目描述
film表
字段 | 说明 |
film_id | 电影id |
title | 电影名称 |
description | 电影描述信息 |
CREATE TABLE IF NOT EXISTS film (
film_id smallint(5) NOT NULL DEFAULT '0',
title varchar(255) NOT NULL,
description text,
PRIMARY KEY (film_id));
category表
字段 | 说明 |
category_id | 电影分类id |
name | 电影分类名称 |
last_update | 电影分类最后更新时间 |
CREATE TABLE category (
category_id tinyint(3) NOT NULL ,
name varchar(25) NOT NULL, `last_update` timestamp,
PRIMARY KEY ( category_id ));
film_category表
字段 | 说明 |
film_id | 电影id |
category_id | 电影分类id |
last_update | 电影id和分类id对应关系的最后更新时间 |
CREATE TABLE film_category (
film_id smallint(5) NOT NULL,
category_id tinyint(3) NOT NULL, `last_update` timestamp);
使用子查询的方式找出属于Action分类的所有电影对应的title,description
输入描述:
[code]无
输出描述:
[code]无
代码:
[code]SELECT f.title,f.description FROM film AS f WHERE f.film_id IN(SELECT fc.film_id FROM film_category AS fc,category AS c WHERE fc.category_id=c.category_id AND c.name='Action')
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第三十题:
题目描述
获取select * from employees对应的执行计划
输入描述:
[code]无
输出描述:
[code]无
代码:
[code]EXPLAIN SELECT * FROM employees
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第三十一题:
题目描述
将employees表的所有员工的last_name和first_name拼接起来作为Name,中间以一个空格区分
CREATE TABLE `employees` ( `emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
输入描述:
[code]无
输出描述:
[code]
Name |
---|
Facello Georgi |
Simmel Bezalel |
Bamford Parto |
Koblick Chirstian |
Maliniak Kyoichi |
Preusig Anneke |
Zielinski Tzvetan |
Kalloufi Saniya |
Peac Sumant |
Piveteau Duangkaew |
Sluis Mary |
代码:
[code]SELECT(last_name||" "||first_name)AS Name FROM employees
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第三十二题:
题目描述
创建一个actor表,包含如下列信息
列表 | 类型 | 是否为NULL | 含义 |
---|---|---|---|
actor_id | smallint(5) | not null | 主键id |
first_name | varchar(45) | not null | 名字 |
last_name | varchar(45) | not null | 姓氏 |
last_update | timestamp | not null | 最后更新时间,默认是系统的当前时间 |
输入描述:
[code]无
代码:
[code]CREATE TABLE actor( actor_id smallint(5) NOT NULL PRIMARY KEY, first_name varchar(45) NOT NULL, last_name varchar(45) NOT NULL, last_update timestamp NOT NULL DEFAULT(datetime('now','localtime')) )
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第三十三题:
题目描述
对于表actor批量插入如下数据
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime('now','localtime')))
actor_id | first_name | last_name | last_update |
---|---|---|---|
1 | PENELOPE | GUINESS | 2006-02-15 12:34:33 |
2 | NICK | WAHLBERG | 2006-02-15 12:34:33 |
输入描述:
[code]无
代码:
[code]INSERT INTO actor VALUES(1,'PENELOPE','GUINESS','2006-02-15 12:34:33'), (2,'NICK','WAHLBERG','2006-02-15 12:34:33')
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第三十四题:
题目描述
对于表actor批量插入如下数据,如果数据已经存在,请忽略,不使用replace操作
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime('now','localtime')))
actor_id | first_name | last_name | last_update |
---|---|---|---|
'3' | 'ED' | 'CHASE' | '2006-02-15 12:34:33' |
输入描述:
[code]无
代码:
[code]INSERT OR IGNORE INTO actor VALUES('3','ED','CHASE','2006-02-15 12:34:33')
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第三十五题:
题目描述
对于如下表actor,其对应的数据为:
actor_id | first_name | last_name | last_update |
---|---|---|---|
1 | PENELOPE | GUINESS | 2006-02-15 12:34:33 |
2 | NICK | WAHLBERG | 2006-02-15 12:34:33 |
创建一个actor_name表,将actor表中的所有first_name以及last_name导入改表。 actor_name表结构如下:
列表 | 类型 | 是否为NULL | 含义 |
---|---|---|---|
first_name | varchar(45) | not null | 名字 |
last_name | varchar(45) | not null | 姓氏 |
输入描述:
[code]无
代码:
[code]CREATE TABLE actor_name AS SELECT first_name ,last_name FROM actor
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第三十六题:
题目描述
针对如下表actor结构创建索引:
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime('now','localtime')))
对first_name创建唯一索引uniq_idx_firstname,对last_name创建普通索引idx_lastname
输入描述:
[code]无
代码:
[code]CREATE UNIQUE INDEX uniq_idx_firstname ON actor(first_name); CREATE INDEX idx_lastname ON actor(last_name)
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第三十七题:
题目描述
针对actor表创建视图actor_name_view,只包含first_name以及last_name两列,并对这两列重新命名,first_name为first_name_v,last_name修改为last_name_v:
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime('now','localtime')))
输入描述:
[code]无
代码:
[code]CREATE VIEW actor_name_view AS SELECT first_name AS first_name_v,last_name AS last_name_v FROM actor
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第三十八题:
题目描述
针对salaries表emp_no字段创建索引idx_emp_no,查询emp_no为10005, 使用强制索引。
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
create index idx_emp_no on salaries(emp_no);
输入描述:
[code]无
代码:
[code]SELECT* FROM salaries INDEXED BY idx_emp_no WHERE emp_no='10005'
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第三十九题:
题目描述
存在actor表,包含如下列信息:
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime('now','localtime')));
现在在last_update后面新增加一列名字为create_date, 类型为datetime, NOT NULL,默认值为'0000 00:00:00'
输入描述:
[code]无
代码:
[code]ALTER TABLE actor ADD COLUMN create_date datetime NOT NULL DEFAULT '0000-00-00 00:00:00'
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第四十题:
题目描述
构造一个触发器audit_log,在向employees_test表中插入一条数据的时候,触发插入相关的数据到audit中。
CREATE TABLE employees_test(
ID INT PRIMARY KEY NOT NULL,
NAME TEXT NOT NULL,
AGE INT NOT NULL,
ADDRESS CHAR(50),
SALARY REAL
);
CREATE TABLE audit(
EMP_no INT NOT NULL,
NAME TEXT NOT NULL
);
输入描述:
[code]无
代码:
[code]CREATE TRIGGER audit_log AFTER INSERT ON employees_test BEGIN INSERT INTO audit(EMP_no,NAME)VALUES(NEW.ID,NEW.NAME); END
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第四十一题:
题目描述
删除emp_no重复的记录,只保留最小的id对应的记录。
CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);
insert into titles_test values ('1', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),
('2', '10002', 'Staff', '1996-08-03', '9999-01-01'),
('3', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'),
('4', '10004', 'Senior Engineer', '1995-12-03', '9999-01-01'),
('5', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),
('6', '10002', 'Staff', '1996-08-03', '9999-01-01'),
('7', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01');
输入描述:
[code]无
代码:
[code]DELETE FROM titles_test WHERE id NOT IN (SELECT MIN(id)FROM titles_test GROUP BY emp_no)
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第四十二题:
题目描述
将所有to_date为9999-01-01的全部更新为NULL,且 from_date更新为2001-01-01。
CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);
insert into titles_test values ('1', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),
('2', '10002', 'Staff', '1996-08-03', '9999-01-01'),
('3', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'),
('4', '10004', 'Senior Engineer', '1995-12-03', '9999-01-01'),
('5', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),
('6', '10002', 'Staff', '1996-08-03', '9999-01-01'),
('7', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01');
输入描述:
[code]无
代码:
[code]UPDATE titles_test SET to_date=NULL,from_date='2001-01-01' WHERE to_date='9999-01-01'
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第四十三题:
题目描述
将id=5以及emp_no=10001的行数据替换成id=5以及emp_no=10005,其他数据保持不变,使用replace实现。
CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);
insert into titles_test values ('1', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),
('2', '10002', 'Staff', '1996-08-03', '9999-01-01'),
('3', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'),
('4', '10004', 'Senior Engineer', '1995-12-03', '9999-01-01'),
('5', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),
('6', '10002', 'Staff', '1996-08-03', '9999-01-01'),
('7', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01');
输入描述:
[code]无
代码:
[code]UPDATE titles_test SET emp_no = REPLACE(emp_no, 10001, 10005) WHERE id = 5 AND emp_no = 10001
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第四十四题:
题目描述
将titles_test表名修改为titles_2017。
CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);
insert into titles_test values ('1', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),
('2', '10002', 'Staff', '1996-08-03', '9999-01-01'),
('3', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'),
('4', '10004', 'Senior Engineer', '1995-12-03', '9999-01-01'),
('5', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),
('6', '10002', 'Staff', '1996-08-03', '9999-01-01'),
('7', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01');
输入描述:
[code]无
代码:
[code]ALTER TABLE titles_test RENAME TO titles_2017
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第四十五题:
题目描述
在audit表上创建外键约束,其emp_no对应employees_test表的主键id。
CREATE TABLE employees_test(
ID INT PRIMARY KEY NOT NULL,
NAME TEXT NOT NULL,
AGE INT NOT NULL,
ADDRESS CHAR(50),
SALARY REAL
);
CREATE TABLE audit(
EMP_no INT NOT NULL,
create_date datetime NOT NULL
);
输入描述:
[code]无
代码:
[code]DROP TABLE audit; CREATE TABLE audit( EMP_no INT NOT NULL, create_date datetime NOT NULL, FOREIGN KEY(EMP_no) REFERENCES employees_test(ID));
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第四十六题:
题目描述
存在如下的视图:
create view emp_v as select * from employees where emp_no >10005;
如何获取emp_v和employees有相同的数据?
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
输出格式:
emp_no | birth_date | first_name | last_name | gender | hire_date |
---|---|---|---|---|---|
10006 | 1953-04-20 | Anneke | Preusig | F | 1989-06-02 |
10007 | 1957-05-23 | Tzvetan | Zielinski | F | 1989-02-10 |
10008 | 1958-02-19 | Saniya | Kalloufi | M | 1994-09-15 |
10009 | 1952-04-19 | Sumant | Peac | F | 1985-02-18 |
10010 | 1963-06-01 | Duangkaew | Piveteau | F | 1989-08-24 |
10011 | 1953-11-07 | Mary | Sluis | F | 1990-01-22 |
代码:
[code]SELECT *FROM emp_v INTERSECT SELECT * FROM employees
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第四十七题:
题目描述
将所有获取奖金的员工当前的薪水增加10%。
create table emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`));
代码:
[code]UPDATE salaries SET salary=salary*1.1 WHERE emp_no IN(SELECT s.emp_no FROM emp_bonus AS eb INNER JOIN salaries AS s ON eb.emp_no=s.emp_no AND s.to_date='9999-01-01')
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第四十八题:
题目描述
针对库中的所有表生成select count(*)对应的SQL语句
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
create table emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输出格式:
cnts |
---|
select count(*) from employees; |
select count(*) from departments; |
select count(*) from dept_emp; |
select count(*) from dept_manager; |
select count(*) from salaries; |
select count(*) from titles; |
select count(*) from emp_bonus; |
代码:
[code]SELECT "select count(*) from "|| name || ";" AS cnts FROM SQLITE_MASTER WHERE type='table'
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第四十九题:
题目描述
将employees表中的所有员工的last_name和first_name通过(')连接起来。
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
输出格式:
name |
---|
Facello'Georgi |
Simmel'Bezalel |
Bamford'Parto |
Koblick'Chirstian |
Maliniak'Kyoichi |
Preusig'Anneke |
Zielinski'Tzvetan |
Kalloufi'Saniya |
Peac'Sumant |
Piveteau'Duangkaew |
Sluis'Mary |
代码:
[code]SELECT last_name||"'"||first_name FROM employees AS name
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第五十题:
题目描述
查找字符串'10,A,B' 中逗号','出现的次数cnt
代码:
[code]SELECT (length('10,A,B')-length(REPLACE('10,A,B',",","")))AS cnt
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第五十一题:
题目描述
获取Employees中的first_name,查询按照first_name最后两个字母,按照升序进行排列
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
输出格式:
first_name |
---|
Chirstian |
Tzvetan |
Bezalel |
Duangkaew |
Georgi |
Kyoichi |
Anneke |
Sumant |
Mary |
Parto |
Saniya |
代码:
[code]SELECT first_name FROM employees ORDER BY SUBSTR(first_name,-2)
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第五十二题:
题目描述
按照dept_no进行汇总,属于同一个部门的emp_no按照逗号进行连接,结果给出dept_no以及连接出的结果employees
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
输出格式:
dept_no | employees |
---|---|
d001 | 10001,10002 |
d002 | 10006 |
d003 | 10005 |
d004 | 10003,10004 |
d005 | 10007,10008,10010 |
d006 | 10009,10010 |
代码:
[code]SELECT dept_no,GROUP_CONCAT(emp_no,',') AS employees FROM dept_emp GROUP BY dept_no
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第五十三题:
题目描述
查找排除当前最大、最小salary之后的员工的平均工资avg_salary。
CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输出格式:
avg_salary |
---|
69462.5555555556 |
代码:
[code]SELECT AVG(salary)AS avg_salary FROM salaries WHERE to_date='9999-01-01' AND salary NOT IN (SELECT MAX(salary)from salaries) AND salary NOT IN (SELECT MIN(salary)from salaries)
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第五十四题:
题目描述
分页查询employees表,每5行一页,返回第2页的数据
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
代码:
[code]SELECT*FROM employees LIMIT 5 OFFSET 5;
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第五十五题:
题目描述
获取所有员工的emp_no、部门编号dept_no以及对应的bonus类型btype和recevied,没有分配具体的员工不显示
CREATE TABLE `dept_emp` ( `emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输出格式:
e.emp_no | dept_no | btype | recevied |
---|---|---|---|
10001 | d001 | 1 | 2010-01-01 |
10002 | d001 | 2 | 2010-10-01 |
10003 | d004 | 3 | 2011-12-03 |
10004 | d004 | 1 | 2010-01-01 |
10005 | d003 | ||
10006 | d002 | ||
10007 | d005 | ||
10008 | d005 | ||
10009 | d006 | ||
10010 | d005 | ||
10010 | d006 |
代码:
[code]SELECT de.emp_no, de.dept_no, eb.btype, eb.recevied FROM dept_emp AS de LEFT JOIN emp_bonus AS eb ON de.emp_no = eb.emp_no;
---------------------------------------------------------------可爱的分界线---------------------------------------------------------------------------------------
第五十六题:
题目描述
使用含有关键字exists查找未分配具体部门的员工的所有信息。
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
输出格式:
emp_no | birth_date | first_name | last_name | gender | hire_date |
---|---|---|---|---|---|
10011 | 1953-11-07 | Mary | Sluis | F | 1990-01-22 |
代码:
[code]SELECT* FROM employees WHERE NOT EXISTS (SELECT emp_no FROM dept_emp WHERE emp_no=employees.emp_no)
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第五十七题:
题目描述
存在如下的视图:
create view emp_v as select * from employees where emp_no >10005;
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
获取employees中的行数据,且这些行也存在于emp_v中。注意不能使用intersect关键字。
输出格式:
emp_no | birth_date | first_name | last_name | gender | hire_date |
---|---|---|---|---|---|
10006 | 1953-04-20 | Anneke | Preusig | F | 1989-06-02 |
10007 | 1957-05-23 | Tzvetan | Zielinski | F | 1989-02-10 |
10008 | 1958-02-19 | Saniya | Kalloufi | M | 1994-09-15 |
10009 | 1952-04-19 | Sumant | Peac | F | 1985-02-18 |
10010 | 1963-06-01 | Duangkaew | Piveteau | F | 1989-08-24 |
10011 | 1953-11-07 | Mary | Sluis | F | 1990-01-22 |
代码:
[code]SELECT* FROM emp_v;
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第五十八题:
题目描述
获取有奖金的员工相关信息。
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
create table emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`));
给出emp_no、first_name、last_name、奖金类型btype、对应的当前薪水情况salary以及奖金金额bonus。 bonus类型btype为1其奖金为薪水salary的10%,btype为2其奖金为薪水的20%,其他类型均为薪水的30%。 当前薪水表示to_date='9999-01-01'
输出格式:
emp_no | first_name | last_name | btype | salary | bonus |
---|---|---|---|---|---|
10001 | Georgi | Facello | 1 | 88958 | 8895.8 |
10002 | Bezalel | Simmel | 2 | 72527 | 14505.4 |
10003 | Parto | Bamford | 3 | 43311 | 12993.3 |
10004 | Chirstian | Koblick | 1 | 74057 | 7405.7 |
代码:
[code]SELECT e.emp_no,e.first_name,e.last_name,eb.btype,s.salary, (CASE eb.btype WHEN 1 THEN s.salary*0.1 WHEN 2 THEN s.salary*0.2 ELSE s.salary*0.3 END)AS bonus FROM employees AS e,emp_bonus AS eb,salaries AS s WHERE s.to_date='9999-01-01' AND e.emp_no=eb.emp_no AND e.emp_no=s.emp_no
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第五十九题:
题目描述
按照salary的累计和running_total,其中running_total为前两个员工的salary累计和,其他以此类推。 具体结果如下Demo展示。。
CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输出格式:
emp_no | salary | running_total |
---|---|---|
10001 | 88958 | 88958 |
10002 | 72527 | 161485 |
10003 | 43311 | 204796 |
10004 | 74057 | 278853 |
10005 | 94692 | 373545 |
10006 | 43311 | 416856 |
10007 | 88070 | 504926 |
10009 | 95409 | 600335 |
10010 | 94409 | 694744 |
10011 | 25828 | 720572 |
代码:
[code]SELECT sb.emp_no,sb.salary,SUM(sa.salary)AS runing_total FROM salaries AS sa,salaries AS sb WHERE sa.emp_no<=sb.emp_no AND sa.to_date='9999-01-01' AND sb.to_date='9999-01-01' GROUP BY sb.emp_no
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
第六十题:
题目描述
对于employees表中,给出奇数行的first_name
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
输出格式:
first_name |
---|
Georgi |
Chirstian |
Anneke |
Tzvetan |
Saniya |
Mary |
代码:
[code]SELECT ea.first_name FROM employees AS ea WHERE (SELECT COUNT(*)FROM employees AS eb WHERE ea.first_name<=eb.first_name)%2==1;
--------------------------------------------------------------可爱的分界线----------------------------------------------------------------------------------------
- 牛客网:数据库SQL实战
- 牛客网:数据库sql实战
- 牛客网-数据库SQL实战
- Gradle实战:执行sql操作hive数据库
- 数据库SQL实战
- 实战SQL 2008 数据库镜像功能
- <实战>oracle 数据库的sql语句中 ‘&’ 符号的处理
- 迁移 SQL Server 数据库到 Azure SQL 实战
- ProxySQL+Mysql实现数据库读写分离实战
- 数据库SQL实战-查找最晚入职员工
- 数据库SQL实战
- 数据库SQL实战
- 【数据库SQL实战】查找薪水涨幅超过15次的员工号emp_no以及其对应的涨幅次数t
- 数据库SQL实战
- 实战 - hive 数据库SQL操作
- 牛客网SQL实战二刷 | Day9
- 【软件自动化测试-QTP实战技能 3】== 数据库检查点手动SQL写法
- 数据库SQL实战
- [数据库SQL实战] 习题及答案
- [牛客数据库SQL实战] 21~30题及个人解答