MySQL----数据表内容操作实战

1、示例

       

　　计算每一个同学的总成绩从大到小排名（显示字段有name，总成绩）

　　　　不加DESC，默认从小到大排序(ASC)

SELECT NAME,SUM(score) AS t FROM u GROUP BY NAME ORDER BY t DESC

　　计算每一个同学的最高成绩（显示字段有name，总成绩）

SELECT NAME,MAX(score) AS t FROM u GROUP BY NAME

2、示例

　　需求，A表关联了B表，如果给A表插入数据

　　就按照正常的逻辑插入数据，但是如果插入的id在B表中找不到，就会报错

INSERT INTO employee(NAME,dept) VALUE("小明",1); //1是B表的id

3、示例

　　关联表的问题，如果from表的时候取了别名，后面只能使用 别名.id

SELECT * FROM realauth AS r  LEFT JOIN logininfo ON r.`applier_id` = logininfo.`id`

4、示例

　　一张表关联多张表

SELECT r.id,ap.id AS ap_id,ap.username,au.id AS au_id,au.username FROM realauth AS r LEFT JOIN logininfo AS ap ON r.`applier_id` = ap.`id` LEFT JOIN logininfo AS au ON r.`auditor_id` = au.`id`

5、示例

　　查出字段不为空和为空的数据

SELECT * FROM userfile WHERE fileType_id IS NOT NULL;
SELECT * FROM userfile WHERE fileType_id IS NULL;

6、示例

    <select id="getClientSchedules" resultMap="CourseScheduleMap">
        SELECT cs.id as scheduleId,ci.courseName,tch.userName teacherName,sv.venueName,sc.campusName,
        cs.weekTimes,cs.beginTimes,cs.endTimes,cs.rebuildLim,cs.genderLim,cs.totalLim,cs.lastLim,
        cs.totalCount,cs.maleScale,cs.maleCount,cs.femaleScale,cs.femaleCount,stime.`shortName` beginTimesShortName,etime.`shortName` endTimesShortName
        FROM (select * from course_schedule where status = 1 and schoolId = #{schoolId}
        <if test="courseId != null">
            and courseId = #{courseId}
        </if>
        <if test="teacherId != null">
            and teacherId = #{teacherId}
        </if>
        <if test="beginTimes != null">
            and beginTimes = #{beginTimes}
        </if>
        <if test="endTimes != null">
            and endTimes = #{endTimes}
        </if>
        <if test="weekTimes != null">
            and weekTimes = #{weekTimes}
        </if>
        <if test="rebuildLim != null and rebuildLim==2">
            and (rebuildLim = 0 or rebuildLim = 2)
        </if>
        <if test="rebuildLim != null and rebuildLim==1">
            and (rebuildLim =0 or rebuildLim = 1)
        </if>
        ) cs
        LEFT JOIN user_info tch ON cs.teacherId= tch.id
        LEFT JOIN course_info ci ON cs.courseId=ci.id
        LEFT JOIN school_venue sv ON cs.venueId = sv.id
        LEFT JOIN school_campus sc ON sv.campusId = sc.id
        LEFT JOIN school_year sy ON cs.yearId = sy.id
        LEFT JOIN school_times_info stime ON stime.`times` = cs.`beginTimes` and stime.status=1 and cs.schoolId= stime.schoolId
        LEFT JOIN school_times_info etime ON etime.`times` = cs.`endTimes` and etime.status=1 and cs.schoolId= etime.schoolId
        LEFT JOIN course_schedule_depart dep ON cs.id= dep.scheduleId
        LEFT JOIN course_schedule_major maj ON cs.id= maj.scheduleId
        LEFT JOIN (select * from user_info where id = #{stuId}) stu ON cs.schoolId= stu.schoolId
        LEFT JOIN school_class scl ON stu.classId= scl.id and scl.status=1
        LEFT JOIN course_class_selection ccs ON ccs.classId = stu.classId

        where sy.isThisYear=1 and (dep.departId is null or dep.departId=stu.departmentId)
        and (maj.majorId is null or maj.majorId=stu.majorId)
        and (ccs.id is null or (cs.weekTimes=ccs.weekTimes and cs.beginTimes>=ccs.beginTimes and ccs.endTimes>=cs.endTimes))
        and ((stu.userSex=0 and  genderLim!=1) or (stu.userSex=1 and  genderLim!=2))
    </select>

7、获取数据库中时间的后几天

语法结构

DATE_SUB(date,INTERVAL expr type)
DATE_ADD(date,INTERVAL expr type)
select * from table where date_sub('2019-04-03',INTERVAL  1 year);

SELECT DATE_ADD(weekBegin,INTERVAL 1 DAY) FROM school_week WHERE WEEK>=1 AND WEEK<=4 AND yearId=22
SELECT DATE_SUB(weekBegin,INTERVAL -1 DAY) FROM school_week WHERE WEEK>=1 AND WEEK<=4 AND yearId=22

比较时间(更多参考:https://www.cnblogs.com/Darkqueen/p/9264087.html)

https://blog.csdn.net/qq_23375733/article/details/88533006

    select
    *
    from attend_record
    where status = 1 and schoolId=#{schoolId} and stuId = #{stuId}  and unix_timestamp(#{signInTime})> unix_timestamp(signInTime)

8、获取薪水表中第二高的薪水

思路1：从大到小排序，取第二

select distinct salary from Employee order by salary desc limit 1,1

思路2：不是最大数的最大数

SELECT max(Salary) SecondHighestSalary
 FROM Employee  
where Salary != (select max(Salary) from Employee ); 

9、分数排名

题目：letcode算法（https://leetcode-cn.com/problems/rank-scores/）

下面这个应该只能支持mysql8.x

SELECT Score,
dense_rank() over(order by Score desc) as 'Rank'
FROM Scores

方式2

select

a.score as Score,

count(DISTINCT b.score) AS Rank # 统计b表符合条件的不重复的分数的数量作为排名

FROM scores a join scores b

where b.score >= a.score # 条件是这个分数不小于我，因为a、b表数据相同，所以排名值最小是1

group by a.id # a表中每个数据都进行排名

order by a.score DESC # 最后按分数（跟排名一样）降序排列

10、查询重复数据

1.having只能用在group by之后，对分组后的结果进行筛选(即使用having的前提条件是分组)。

select Email from Person group by Email having count(Email) > 1