Question

我有一张“Songs”，“Songs_Tags”（与歌曲相关的歌曲）和“Songs_Votes”（与布尔喜欢/不喜欢的歌曲相关）的表格。

我需要使用其标签的GROUP_CONCAT（）以及喜欢的数量（真实）和不喜欢（虚假）来检索歌曲。

我的查询是这样的：

SELECT
    s.*,
    GROUP_CONCAT(st.id_tag) AS tags_ids,
    COUNT(CASE WHEN v.vote=1 THEN 1 ELSE NULL END) as votesUp,
    COUNT(CASE WHEN v.vote=0 THEN 1 ELSE NULL END) as votesDown,
FROM Songs s
    LEFT JOIN Songs_Tags st ON (s.id = st.id_song)
    LEFT JOIN Votes v ON (s.id=v.id_song)
GROUP BY s.id
ORDER BY id DESC

问题是，当一首歌有超过1个标签时，它会被多次返回，所以当我执行COUNT（）时，它会返回更多结果。

我能想到的最好的解决方案是，在GROUP BY之后是否可以进行最后的LEFT JOIN（所以现在每首歌只有一个条目）。然后我需要另一个GROUP BY m.id。

有没有办法实现这一目标？我需要使用子查询吗？

Answer 1

到目前为止已经有了一些很好的答案，但我会采用与你原来描述的方法略有不同的方法

SELECT
    songsWithTags.*,
    COALESCE(SUM(v.vote),0) AS votesUp,
    COALESCE(SUM(1-v.vote),0) AS votesDown
FROM (
    SELECT
        s.*,
        COLLATE(GROUP_CONCAT(st.id_tag),'') AS tags_ids
    FROM Songs s
    LEFT JOIN Songs_Tags st
        ON st.id_song = s.id
    GROUP BY s.id
) AS songsWithTags
LEFT JOIN Votes v
ON songsWithTags.id = v.id_song

GROUP BY songsWithTags.id DESC

在此，子查询负责将带有标签的歌曲整理为每首歌曲1行。然后将其加入投票。我还选择简单地总结v.votes列，因为你已经指出它是1或0，因此SUM（v.votes）将加起来1 + 1 + 1 + 0 + 0 = 3中的3个是upvotes，而SUM（1-v.vote）将总和0 + 0 + 0 + 1 + 1 = 2中的5个是downvotes。

如果你有一个带有列（id_song，vote）的投票索引，那么该索引将用于此，所以它甚至不会命中表。同样，如果您在Songs_Tags上有一个带有（id_song，id_tag）的索引，那么该表将不会被查询命中。

编辑使用计数

添加解决方案

SELECT
    songsWithTags.*,
    COUNT(CASE WHEN v.vote=1 THEN 1 END) as votesUp,
    COUNT(CASE WHEN v.vote=0 THEN 1 END) as votesDown
FROM (
    SELECT
        s.*,
        COLLATE(GROUP_CONCAT(st.id_tag),'') AS tags_ids
    FROM Songs s
    LEFT JOIN Songs_Tags st
        ON st.id_song = s.id
    GROUP BY s.id
) AS songsWithTags
LEFT JOIN Votes v
ON songsWithTags.id = v.id_song

GROUP BY songsWithTags.id DESC

Answer 2

试试这个：

SELECT
    s.*,
    GROUP_CONCAT(DISTINCT st.id_tag) AS tags_ids,
    COUNT(DISTINCT CASE WHEN v.vote=1 THEN id_vote ELSE NULL END) AS votesUp,
    COUNT(DISTINCT CASE WHEN v.vote=0 THEN id_vote ELSE NULL END) AS votesDown
FROM Songs s
    LEFT JOIN Songs_Tags st ON (s.id = st.id_song)
    LEFT JOIN Votes v ON (s.id=v.id_song)
GROUP BY s.id
ORDER BY id DESC

Answer 3

您的代码会产生迷你笛卡尔积，因为您在1-to-many关系中进行了两次连接，而1表位于两个连接的同一侧。

使用分组转换为2个子查询，然后加入：

SELECT
    s.*,
    COALESCE(st.tags_ids, '') AS tags_ids,
    COALESCE(v.votesUp, 0)    AS votesUp,
    COALESCE(v.votesDown, 0)  AS votesDown
FROM 
        Songs AS s
    LEFT JOIN 
        ( SELECT 
              id_song,
              GROUP_CONCAT(id_tag) AS tags_ids
          FROM Songs_Tags 
          GROUP BY id_song
        ) AS st
      ON s.id = st.id_song
    LEFT JOIN 
        ( SELECT
              id_song,
              COUNT(CASE WHEN v.vote=1 THEN id_vote END) AS votesUp,
              COUNT(CASE WHEN v.vote=0 THEN id_vote END) AS votesDown
          FROM Votes 
          GROUP BY id_song
        ) AS v 
      ON s.id = v.id_song
ORDER BY s.id DESC

在GROUP BY之后左转加入？

3 个答案: