我有一个视频表
表格:' id',' video_key,'周'
值 :( 1,12345,2016-01-03),(2,23456,2016-01-03),(3,3 3456,2015-01-03)等。 ..
周'将永远是一些星期日日期,每个星期'可以有任意数量的视频。
我试图提出一个查询按周排序视频的问题' DESC,并且具有降序(递减?)排名。我可以使用以下内容按照递增的顺序获取视频 - 但这不是我需要的结果。
SELECT tt.*, rank FROM ( SELECT t.*, @week:=
CASE WHEN @week = week
THEN @rank:=@rank +1
ELSE @rank:=1
END rank,
@week:=t.week
FROM video_table t ,
(SELECT @rank:=0,@week:=0) r
ORDER BY week DESC, id ASC
) tt
返回类似
的内容[id,video_key,week,rank,...] 1,12345,2016-01-03,1,.. 2,23456,2016-01-03,2,.. 3,3456,2016-01-03,3,....
但实际上我需要它
[id,video_key,week,rank,...] 3,3456,2016-01-03,1,.. 2,23456,2016-01-03,2,.. 1,12345,2015-01-03,3
我的想法是我需要做一个子查询来获取每周视频的计数(*),然后执行以下操作:
CASE WHEN @week = week
THEN @weekly_count_of_videos:=@weekly_count_of_videos -1
//....etc
我希望这一切都有意义....感谢任何提示,或指示实现这一目标。
基于Gordon Linoff回答的更新查询
SELECT t3.* FROM (SELECT t2.* FROM ( SELECT t1.*, @week:= CASE WHEN @week = week THEN @rank:=@rank +1 ELSE @rank:=1 END rank, @week:=t1.week FROM video_table t1 , (SELECT @rank:=0,@week:=0) r ORDER BY week DESC, id ASC ) t2) t3 order by week DESC, rank desc
答案 0 :(得分:1)
只需将查询作为子查询并按外部查询:
select t.*
from (<your query>
) t
order by rank desc;