sql语句使用记录

根据用户链接上数据库命令：mysql -u root -p

然后输入密码

[root@ovp-e5420-002 wuxianji]# mysql -u wuxianji_root -p

Enter password: 

查询存在重复的字段（mysql）

select nickname from user where school_id="swiftv" group by nickname having count(*)>1;

SELECT user_id FROM statistics_duration_cu GROUP BY user_id HAVING count(`user_id`) >1;

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按月份统计的查询，每月注册人数：

select DATE_FORMAT(date_added,'%Y-%m') as month,count(*) as shumu from user  

where school_id='swiftv' and DATE_FORMAT(date_added,'%Y')=2014 group by month order by month;

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链接查询：

左连接的两种条件过滤查询方式：

-- 链接查询，后再过滤
SELECT uw.ITEM_USER_ID,uw.ITEM_WORK_ID,wm.* from tlk_uid_wid as uw
LEFT JOIN tlk_workrecordmonth as wm on uw.ITEM_WORK_ID = wm.ITEM_WORK_ID
where uw.ITEM_USER_ID="11e6-1ca7-fa6a5f88-9915-490ea07f921f";

-- 过滤后，在链接查询
SELECT uw.ITEM_USER_ID,uw.ITEM_WORK_ID,wm.* from
(select * from tlk_uid_wid where ITEM_USER_ID="11e6-1ca7-fa6a5f88-9915-490ea07f921f") as uw
LEFT JOIN tlk_workrecordmonth as wm on uw.ITEM_WORK_ID = wm.ITEM_WORK_ID;
左链接：(代码格式分段注意空格)，注意条件位置

select a.user_id, a.school_id,a.nickname, a.avatar, c.user_id, c.title, c.category_id, c.is_free
from user as a
LEFT JOIN course as c on a.user_id=c.user_id
where a.school_id="test";

全表在左，用左链接查询， 如果全表在右侧可以用右链接查询 （user表为全表）

select a.user_id, a.school_id,a.nickname, a.avatar, c.user_id, c.title, c.category_id, c.is_free
from user as a
LEFT JOIN course as c on a.user_id=c.user_id
and c.is_free="Y" //查询course表的条件
where a.school_id="test";

条件与上面语句位置不同的左连接查询，结果与上面不同，但与下面的右连接查询结果相同

select a.user_id, a.school_id,a.nickname, a.avatar, c.user_id, c.title, c.category_id, c.is_free
from user as a
LEFT JOIN course as c on a.user_id=c.user_id
where a.school_id="test" and c.is_free="Y";

右链接：（course为全表）

select a.user_id, a.school_id,a.nickname, a.avatar, c.user_id, c.title, c.category_id, c.is_free
from user as a
RIGHT JOIN course as c on a.user_id=c.user_id  //链接条件
and c.is_free="Y" //course查询条件
where a.school_id="test";

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三个表的连接查询：

select a.user_id, is_instructor,a.school_id,a.nickname, a.avatar, c.user_id, c.title, c.category_id, c.is_free, t.json_content, t.course_id
from user as a
LEFT JOIN course as c on a.user_id=c.user_id
LEFT JOIN course_content as t on  c.course_id=t.course_id
and c.is_free="Y"
where a.school_id="test" and a.is_instructor="Y";

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以一个表不存在的id为条件去删除另外一个表的数据

delete from A

where id not in (select id from B);

DELETE from question 

WHERE course_id not in (select course_id from course);

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三表关联查询，最新回答（答案表-----问题表-----课程表）mysql语句

 SELECT an.*, q.questionTitle, c.courseType, c.courseTitle, c.courseSubtitle, c.coverImage
 FROM (
         SELECT * 
         FROM answer WHERE school_id="swiftv" 
      ) AS an,

        (
         SELECT question_id AS questionId, title AS questionTitle 
         FROM question 
         WHERE school_id = "swiftv" 
      ) AS q,
      (
         SELECT course_id AS courseId, course_type AS courseType, title AS courseTitle, subtitle AS courseSubtitle, cover_image AS coverImage
         FROM course
         WHERE school_id = "swiftv" 
      ) AS c
 WHERE an.question_id = q.questionId AND an.course_id = c.courseId 
 ORDER BY an.date_added DESC 
 LIMIT 0,5;

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查看一天的记录

select * from statistics_duration where date_end BETWEEN '2014-09-18 00:00:00' AND '2014-09-19 00:00:00' GROUP BY user_id;

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查看当天的数据

select * from statistics_duration_cu where to_days(`date`) = to_days(now()) and user_id=#{userId,jdbcType=VARCHAR} and course_id=#{courseId,jdbcType=VARCHAR} and lecture_id=#{lectureId,jdbcType=VARCHAR} and school_id=#{schoolId,jdbcType=VARCHAR};

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mysql中把查询出来的结果在进行查询

/*root界面，所有时长统计*/

select t1.schoolId,t1.sname,t1.sdomain,t1.dDuration,t1.dAmount,t1.dCount,sum(sd.duration)*60 as sdDuration from

(SELECT dl.school_id AS schoolId,sum(dl.duration) AS dDuration,sum(dl.amount) as dAmount,count(dl.school_id) as dCount,sl.name as sname,sl.domain as sdomain from deposit_log as dl 

 LEFT JOIN school as sl on sl.school_id=dl.school_id 

 where dl.status='Y' GROUP BY dl.school_id) as t1 LEFT JOIN statistics_duration_cu as sd on t1.schoolId=sd.school_id GROUP BY t1.schoolId;

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1。表结构相同的表，且在同一数据库（如，table1,table2)

Sql ：insert into table1 select * from table2 (完全复制)

           insert into table1 select distinct * from table2(不复制重复纪录）

          insert into table1 select top 5 * from table2 (前五条纪录)

2。   不在同一数据库中（如，db1 table1,db2 table2)

sql:    insert into db1..table1 select * from db2..table2 (完全复制)

           insert into db1..table1 select distinct * from db2table2(不复制重复纪录）

          insert into tdb1..able1 select top 5 * from   db2table2 (前五条纪录)

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