Question

我很难过，摸不着头脑。它一定很简单，但我没有看到它。

假设我有四张桌子：

video = id
hastag = id, tag_id, video_id
hasteam = id, team_id, video_id
hasidol = id, idol_id, video_id

此数据集（仅作为示例）：

video = (1), (2), (3)
hastag = (1, 1, 1), (2, 1, 2), (3, 2, 3)
hasteam = (1, 1, 1), (2, 1, 3), (3, 2, 2)
hasidol = (1, 1, 3)

这个查询：

SELECT v.id ,
COUNT(vhtag.id),
COUNT(vhteam.id),
COUNT(vhidol.id)
FROM video v
LEFT JOIN hastag vhtag ON vhtag.video_id = v.id
LEFT JOIN hasteam vhteam ON vhteam.video_id = v.id
LEFT JOIN hasidol vhidol ON vhidol.video_id = v.id

WHERE
v.id <> 1
AND
(
  vhtag.tag_id IN (SELECT htt2.tag_id FROM hastag htt2 WHERE video_id = 1)
  OR
  vhteam.team_id IN (SELECT htt3.team_id FROM hasteam htt3 WHERE video_id = 1)
  OR
  vhidol.idol_id IN (SELECT htt4.idol_id FROM hasidol htt4 WHERE video_id = 1)
)
GROUP BY v.id

它给出了与“WHERE”子句不对应的“has”行数。例如，如果视频行只有一个视频ID为1的团队和一个完全不相关的标签，那么当它应该说“普通标签”时，它会给我“普通标签数量：1，普通团队数量：1” count：0（因为它是一个不相关的标签），常见的团队数：1“。

现在，只要我将查询限制为只有一个“has”表，就像这样：

SELECT v.id ,
COUNT(vhtag.id)
FROM video v
LEFT JOIN hastag vhtag ON vhtag.video_id = v.id

WHERE
v.id <> 1
AND
(
  vhtag.tag_id IN (SELECT htt2.tag_id FROM hastag htt2 WHERE video_id = 1)
)
GROUP BY v.id

然后它确实有效，但问题是当我尝试在查询中放入多个“has”表时。我尝试过使用HAVING，但它无法识别“vhtag.tag_id”列。我显然在这里做错了，有人可以帮帮我吗？

编辑：

这种作品：

LEFT JOIN hastag vhtag ON vhtag.video_id = v.id AND vhtag.tag_id IN (SELECT htt2.tag_id FROM hastag htt2 WHERE video_id = 1)
LEFT JOIN hasteam vhteam ON vhteam.video_id = v.id AND vhteam.team_id IN (SELECT htt3.team_id FROM hasteam htt3 WHERE video_id = 1)
LEFT JOIN hasidol vhidol ON vhidol.video_id = v.id AND vhidol.idol_id IN (SELECT htt4.idol_id FROM hasidol htt4 WHERE video_id = 1)

我也可以在Doctrine中使用它（我是愚蠢的，忘了WITH）。这是最佳方式吗？

Answer 1

尝试简化查询，例如以这种方式 -

SELECT
  v.id,
  COUNT(vhtag.id),
  COUNT(vhteam.id),
  COUNT(vhidol.id)
FROM
  video v
LEFT JOIN (SELECT * FROM hastag WHERE video_id = 1) vhtag
  ON vhtag.video_id = v.id
LEFT JOIN (SELECT * FROM hasteam WHERE video_id = 1) vhteam
  ON vhteam.video_id = v.id
LEFT JOIN (SELECT * FROM hasidol WHERE video_id = 1) vhidol
  ON vhidol.video_id = v.id
WHERE
  v.id <> 1
GROUP BY
  v.id

它能给出正确的结果吗？

输出：

+------+-----------------+------------------+------------------+
| id   | COUNT(vhtag.id) | COUNT(vhteam.id) | COUNT(vhidol.id) |
+------+-----------------+------------------+------------------+
|    2 |               0 |                0 |                0 |
|    3 |               0 |                0 |                0 |
+------+-----------------+------------------+------------------+

Answer 2

您遇到的问题是，您的联接会导致每个团队中的交叉联接。

最简单的方法是统计不同：

SELECT v.id , COUNT(distinct vhtag.id), COUNT(distinct vhteam.id),
       COUNT(distinct vhidol.id)
FROM video v LEFT JOIN
     hastag vhtag
     ON vhtag.video_id = v.id
     LEFT JOIN hasteam vhteam
     ON vhteam.video_id = v.id
     LEFT JOIN hasidol vhidol
     ON vhidol.video_id = v.id

真正的解决方案是分别聚合每个值，然后将结果连接在一起。

分组，计数和在哪里

2 个答案: