从这个查询:查询GROUP BY和ORDER BY
SELECT
v.idutente AS idutente,
vl.idvideo AS idvideo,
(vl.likeY-vl.unlikeY) AS sott
FROM video_likeunlike AS vl
LEFT OUTER JOIN video AS v
ON vl.idvideo = v.ID
WHERE status = '1'
我从表
idutente idvideo likeY unlikeY sott
------------------------------------------------
1 70 2 5 -3
2 81 6 10 -4
1 52 200 198 2
1 54 30 2 28
2 67 20 5 15
3 22 12 0 12
这个数据,我需要这样的结果,有可能吗?
idutente likeY unlikeY sott
------------------------------------------------
1 232 205 27
2 26 15 11
3 12 0 12
我尝试此查询
SELECT
v.idutente AS idutente,
vl.idvideo AS idvideo,
vl.likeY as likeY,
vl.unlikeY as unlikeY,
(vl.likeY-vl.unlikeY) AS sott
FROM video_likeunlike AS vl
LEFT OUTER JOIN video AS v
ON vl.idvideo = v.ID
WHERE status = '1'
GROUP BY v.idutente
ORDER BY sott DESC
您正在使用2台在您尝试查询,但你仅列1中的问题表。 – wast
请参阅:[为什么我应该为我认为是非常简单的SQL查询提供一个MCVE?](https://meta.stackoverflow.com/questions/333952/why-should-i-provide-an-mcve -for - 什么 - 似乎对我将要-A-极简单的SQL查询) – Strawberry