8.View the Exhibit and examine the structure of the CUSTOMERS table.
2016-01-19 11:24
375 查看
8.View the Exhibit and examine the structure of the CUSTOMERS table.
Which two tasks would require subqueries or joins to be executed in a single statement? (Choose two.)
A.listing of customers who do not have a credit limit and were born before 1980
B.finding the number of customers, in each city, whose marital status is 'married'
C.finding the average credit limit of male customers residing in 'Tokyo' or 'Sydney'
D.listing of those customers whose credit limit is the same as the credit limit of customers residing in the city 'Tokyo'
E.finding the number of customers, in each city, whose credit limit is more than the average credit limit of all the customers
答案:DE
解析:题目问的是ABCD哪个需要子查询
A:select * from customers where nvl(cust_create_limit,0)=0 and cust_year_of_birth<1980
B:select cust_city,count(*) from customers where cust_marital_status = 'married' group by cust_city
C:select avg(cust_create_limit) from customers where cust_gender=0 and cust_city in ('Tokyo','Sydney')
D:select * from customers where cust_create_limit in (select cust_create_limit from customers where cust_city='Tokyo'
E:select cust_city,count(*) from customers where cust_create_limit >(select avg(cust_create_limit) from customers) group by cust_city
因此答案:DE
Which two tasks would require subqueries or joins to be executed in a single statement? (Choose two.)
A.listing of customers who do not have a credit limit and were born before 1980
B.finding the number of customers, in each city, whose marital status is 'married'
C.finding the average credit limit of male customers residing in 'Tokyo' or 'Sydney'
D.listing of those customers whose credit limit is the same as the credit limit of customers residing in the city 'Tokyo'
E.finding the number of customers, in each city, whose credit limit is more than the average credit limit of all the customers
答案:DE
解析:题目问的是ABCD哪个需要子查询
A:select * from customers where nvl(cust_create_limit,0)=0 and cust_year_of_birth<1980
B:select cust_city,count(*) from customers where cust_marital_status = 'married' group by cust_city
C:select avg(cust_create_limit) from customers where cust_gender=0 and cust_city in ('Tokyo','Sydney')
D:select * from customers where cust_create_limit in (select cust_create_limit from customers where cust_city='Tokyo'
E:select cust_city,count(*) from customers where cust_create_limit >(select avg(cust_create_limit) from customers) group by cust_city
因此答案:DE
相关文章推荐
- PHP扩展迁移为兼容PHP7记录
- PHP 获取图片中的器材信息
- 位运算小结(按位与、按位或、按位异或、取反、左移、右移)
- 用 Webgoat 撬动地球,看安全测试的引路石!
- stsiLdetroSowTegreM.21
- VS2013自带报表+打印功能
- iPhone 6 Screens Demystified
- python中numpy库matrix和array的融合使用
- Android Volley完全解析(一),初识Volley的基本用法
- 拦截PHP各种异常和错误,发生致命错误时进行报警,万事防患于未然
- 生成唯一订单号的算法
- 数据库连接池介绍
- JAVA线程join用法
- Linux系统中nc命令的基本用法掌握
- linux下导入、导出mysql数据库命令
- c++ 封装 (上)
- 《iOS Human Interface Guidelines》——In-App Purchase
- 监听静音开关
- JavaScript基础:数据类型的中的那些少见多怪
- 打印org.eclipse.xsd.XSDSchema对象