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mysql 根据总分排名

时间:2019-04-15 15:04:41      阅读:133      评论:0      收藏:0      [点我收藏+]
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

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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; 

 

mysql 根据总分排名

原文:https://www.cnblogs.com/NCL--/p/10710574.html

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