mysql 根据总分排名

最新推荐文章于 2024-05-24 22:37:06 发布

转载最新推荐文章于 2024-05-24 22:37:06 发布 · 278 阅读

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原文链接：http://www.cnblogs.com/NCL--/p/10710574.html

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#数据库

mysql 根据总分排名

SELECT
    t.*,
    @rank := @rank + 1 AS rank 
FROM
    ( SELECT @rank := 0 ) r,
    (
SELECT
    tas.id,
    tas.teacher_id,
    tas.average,
    tas.remark,
    tas.create_time,
    tas.course_id,
    su.user_name,
    ti.post,
    ti.SUBJECT 
FROM
    teacher_average_score tas
    LEFT JOIN sys_user su ON tas.teacher_id = su.user_id
    LEFT JOIN teacher_info ti ON tas.teacher_id = ti.user_id 
ORDER BY
    tas.average DESC 
    ) AS t

mysql数据库查询表中相邻数据的差值

select a.time ,a.sum - b.sum sum,a.time,b.time
from  
(select @arownum:=@arownum+1 rownum,sum ,time from summary_hour,(select @arownum:=0) t where energy_type= 1 order by time) a,  
(select @brownum:=@brownum+1 rownum ,sum,time from summary_hour,(select @brownum:=1) t where energy_type= 1 ORDEr by time) b  
where a.rownum = b.rownum;