Question

来自此查询：

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'

我有来自表

的数据

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

我需要这个结果，可能吗？

idutente   likeY      unlikeY    sott 
------------------------------------------------
1           232         205        27
2           26          15         11
3           12           0         12

我尝试使用此查询

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

Answer 1

请对选择列使用聚合SUM（）操作。查询可以修改为：

SELECT v.idutente AS idutente,
       vl.idvideo AS idvideo,
       SUM(vl.likeY) AS likeY,
       SUM(vl.unlikeY) AS unlikeY,
       SUM(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

Answer 2

看起来你想使用sql sum()函数

SELECT 
  v.idutente AS idutente, 
  sum(vl.likeY) as likeY, 
  sum(vl.unlikeY) as unlikeY, 
  sum(vl.likeY)-sum(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

Answer 3

SELECT idutente, 
  SUM(likeY) likeY, 
  SUM(unlikeY) unlikeY, 
  SUM(sott) sott
FROM mytable 
GROUP BY idutente
ORDER BY sott DESC

在SQL Fiddle中试用。

Answer 4

将您的第一个查询用作子查询，然后执行分组。

<强>查询

SELECT t.idutente, SUM(t.likeY) as likeY, SUM(t.unlikeY) as unlikeY from(
    SELECT 
      v.idutente AS idutente, 
      vl.idvideo AS idvideo, 
      vl.likeY,
      vl.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'
) t
GROUP BY t.idutente
ORDER t.sott desc;

查询分组依据和顺序

4 个答案: