Question

我希望通过特定的用户ID以及特定网页ID的所有喜欢和喜欢获得喜欢的结果。

我的结构如下：

`pages`: (id, title)
`pages_likes`: (id, page_id, uid, status)

如果status为-1，则表示不喜欢特定网页，如果它是1，则表示不喜欢。

所以，为了得到所有喜欢这是我的疑问：

SELECT COUNT(id) FROM pages_likes WHERE status = '1'

但是现在我也想知道如果用户ID 3例如喜欢

这个页面

SELECT COUNT(id) FROM pages_likes WHERE status = '1' AND uid='3'

如何在一个查询中实现这两个目标？我想在SELECT语句之后必须改变一些东西吗？

Answer 1

我认为这样可行：

SELECT '' user, COUNT(id) likes 
FROM pages_likes 
WHERE status = '1' 
GROUP BY 1

UNION ALL

SELECT uid user, COUNT(id) likes 
FROM pages_likes 
WHERE status = '1' AND uid='3'
GROUP BY 1

Answer 2

如果要在单个查询中执行此操作，请使用条件聚合：

SELECT SUM(CASE WHEN uid = '3' THEN 1 ELSE 0 END) AS threeLikes,
       SUM(CASE WHEN uid <> '3' THEN 1 ELSE 0 END) AS otherLikes
FROM pages_likes
WHERE status = '1'

另一种选择是使用UNION，参见@bernie给出的答案

<强>更新

如果您希望页面中的和不喜欢同一个查询，可以尝试：

SELECT SUM(CASE WHEN uid = '3'  AND status = '1' THEN 1 ELSE 0 END) AS threeLikes,
       SUM(CASE WHEN uid <> '3' AND status = '1' THEN 1 ELSE 0 END) AS otherLikes
       SUM(CASE WHEN uid = '3'  AND status = '0' THEN 1 ELSE 0 END) AS threeDislikes,
       SUM(CASE WHEN uid <> '3' AND status = '0' THEN 1 ELSE 0 END) AS otherDisikes
FROM pages_likes

Answer 3

我会这样做：

SELECT COUNT(*) AS total_likes, SUM(uid=3) AS uid3_likes
FROM pages_likes
WHERE status=1 AND page_id=1234

重新评论：

这是一个显示总喜欢和完全不喜欢的例子。它类似于@TimBiegeleisen的答案。

SELECT SUM(status=1) AS total_likes, 
       SUM(CASE WHEN uid=3 AND status=1 THEN 1 END) AS uid3_likes
       SUM(CASE WHEN uid=3 AND status=-1 THEN 1 END) AS uid3_dislikes
FROM pages_likes
WHERE page_id=1234

使用一个查询获取两个结果

3 个答案: