我正在尝试计算多个表的结果。我有图片,图片有评论,喜欢和不喜欢。我想将所有这些合并到一个表查询中,我将在其中显示所有照片并计算对该照片的喜欢、评论和不喜欢。
我的照片表是这样的:
photo_id owner_id album_id image_type photo_name photo_ext photo_size photo_type photo_description date_uploaded date_midified photo_guid bg_x_position bg_y_position
-------- -------- -------- ---------- ------------------------------------ --------- ---------- ---------- ----------------- ------------------- ------------------- ------------------------------------ ------------- ---------------
2 1 5 0 27bda14cb0e30efb0eeef5688e6c0ef9.jpg .jpg 42.6 jpeg (NULL) 2016-02-09 15:45:20 2016-02-09 15:45:20 E8CFF5F0-F8FA-8AD7-E3B5-63EAFE81E1B6 (NULL) (NULL)
3 1 5 0 8077f0d2104c35c8e612e24834f82ace.jpg .jpg 34.52 jpeg (NULL) 2016-02-09 15:49:44 2016-02-09 15:49:44 09B25F5C-0A9D-0EBC-DCE9-EE24E8ED4B6D (NULL) (NULL)
4 1 5 0 6c28b264470e7a7f2829ea5b7290cbba.jpg .jpg 85.65 jpeg (NULL) 2016-02-09 15:49:56 2016-02-09 15:49:56 9A5EF85E-F691-F42E-C20C-BCDC765BFA1B (NULL) (NULL)
点赞表:
like_id item_id account_id rate time host
------- ------- ---------- ------ ------------------- --------
308 2 1 1 2016-03-18 13:45:16 (NULL)
309 3 1 2 2016-03-18 13:45:33 (NULL)
310 2 7 1 2016-03-18 14:23:49 (NULL)
评论表:
comment_id item_id content account_id time
------- ------- ---------- ------------ -------------------
308 262 Test comment 1 2016-03-18 13:45:16
所以我想要的结果是这样的:
photo_id owner_id album_id image_type photo_name photo_ext photo_size photo_type photo_description date_uploaded date_midified photo_guid bg_x_position bg_y_position likes dislikes comments
-------- -------- -------- ---------- ------------------------------- --------- ---------- ---------- ----------------- ------------------- ------------------- ---- ------ ------ ------------------------------------ ------------- --------------
所以我想在现有表格上添加 3 个新单元格(喜欢、不喜欢和评论)计数。我的查询工作正常,但评论数不正确,因为他显示所有图片都有 1
评论,但没有评论。
SELECT * FROM (SELECT p.photo_id, p.photo_name,
IFNULL(SUM(l.rate = 1), 0) AS likes,
IFNULL(SUM(l.rate = 2), 0) AS dislikes
FROM pb_account_photos AS p
LEFT JOIN pb_account_likes AS l ON l.item_id = p.photo_id
WHERE p.owner_id = 1 GROUP BY p.photo_id) AS LIKES,
(SELECT COUNT(*) AS comments, p.photo_id
FROM pb_account_photos AS p
LEFT JOIN pb_account_comments AS c ON c.item_id = p.photo_id
WHERE p.owner_id = 1 GROUP BY p.photo_id) AS COMM WHERE COMM.photo_id=LIKES.photo_id;
最佳答案
问题是您在 LEFT JOIN
之后进行计数。在这种情况下,COUNT(*)
总是返回一个至少为 1 的值,因为它正在计算行数。您需要计算匹配项,因此计算第二个表中的一列:
SELECT likes.*, comm.comments
FROM (SELECT p.photo_id, p.photo_name,
COALESCE(SUM(l.rate = 1), 0) AS likes,
COALESCE(SUM(l.rate = 2), 0) AS dislikes
FROM pb_account_photos p LEFT JOIN
pb_account_likes l
ON l.item_id = p.photo_id
WHERE p.owner_id = 1
GROUP BY p.photo_id
) likes JOIN
(SELECT COUNT(C.item_id) AS comments, p.photo_id
FROM pb_account_photos p LEFT JOIN
pb_account_comments c
ON c.item_id = p.photo_id
WHERE p.owner_id = 1
GROUP BY p.photo_id
) comm
ON comm.photo_id = likes.photo_id;
关于mysql - 从多个表中计数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36474002/