Question

我目前有一个查询，它返回存在的每个帖子的列表，其评论（连接）以及它有多少喜欢。

我想添加一个列，指示用户是否已赞成/喜欢此特定帖子。相关表格的结构如下：

喜欢（id，elementID，googleID）
元素（id，googleID，时间，标题，正文，类型）
评论（id，elementID，googleID，正文）

那么我将如何浏览喜欢的内容，确定其中一个是否具有用户的ID（通过Node.JS动态调入查询），然后创建一个指示简单的真或假的列？ / p>

SELECT E.id, E.time, E.title, E.body, E.type, C.comments, E.googleID, L.likeCount
FROM elements E
LEFT JOIN(
    SELECT elementID, GROUP_CONCAT(body SEPARATOR '|-|') AS comments
    FROM comments
    GROUP BY elementID
) C on C.elementID = E.id
LEFT JOIN (
    SELECT elementID, COUNT(id) AS likeCount
    FROM likes
    GROUP BY elementID
) L ON L.elementID = E.id;

Answer 1

针对LEFT JOIN添加likes，将用户的googleID传递到加入的ON子句中。如果来自该连接的值为非null，则用户已喜欢该帖子。如果为null，则用户不喜欢它。

请注意，您的第一个子查询连接是不必要的。这可以直接在SELECT列表中进行处理，并将所有其他所选列添加到GROUP BY子句中（实际上，由于其默认行为，在MySQL中甚至不需要这样做）按照宽容分组，但是因为其他RDBMS需要它，所以是一个好习惯。）

SELECT
  E.id,
  E.time,
  E.title,
  E.body,
  E.type,
  -- This can be moved to the outer SELECT
  GROUP_CONCAT(C.body SEPARATOR '|-|') AS comments,
  E.googleID,
  L.likeCount,
  -- Conditional statement to match non-nulls in the
  -- left joined likes table, indicating whether the
  -- user has liked the post
  CASE
    -- has a like, return true
    WHEN Luser.id IS NOT NULL THEN true
    -- does not have a like, return false
    ELSE false
  END AS userLiked
FROM elements E
-- Convert this to a plain left join instead of subquery
LEFT JOIN comments C ON C.elementID = E.id
-- Left join against likes, strictly matching the user id
LEFT JOIN likes Luser
  ON Luser.elementID = E.id
  -- Pass the user id into the join condition.
  AND Luser.googleID = <the user's id>
LEFT JOIN (
    SELECT elementID, COUNT(id) AS likeCount
    FROM likes
    GROUP BY elementID
) L ON L.elementID = E.id;
-- Put the other columns into GROUP BY
-- instead of another left join subquery
GROUP BY
  E.id,
  -- MySQL will let you get by without the rest of these
  -- but it is good practice to use them. Other RDBMS would
  -- not allow them to be omitted...
  E.time,
  E.title,
  E.body,
  E.type,
  E.googleID
  likeCount,
  userLiked

如果您希望返回使用GROUP_CONCAT()的子查询联接，您可以恢复使用并删除外部GROUP BY，但它可能没有此方法那么快。

Answer 2

添加条件SUM这将是0或>0，以防用户喜欢它。

SELECT 
     E.id, E.time, E.title, E.body, E.type, 
     C.comments, E.googleID, L.likeCount, L.user_likeit
FROM elements E
LEFT JOIN(
    SELECT elementID, GROUP_CONCAT(body SEPARATOR '|-|') AS comments
    FROM comments
    GROUP BY elementID
) C on C.elementID = E.id
LEFT JOIN (
    SELECT elementID, COUNT(id) AS likeCount, SUM( CASE 
                                                     WHEN googleID =@userID THEN 1
                                                     ELSE 0
                                                   END
                                                 ) as user_likeit
    FROM likes
    GROUP BY elementID
) L ON L.elementID = E.id;

我谷歌并发现Conditional Count/Sum in Select看起来也可以工作

sum(googleID = @userID)

如何在MySQL查询中包含用户？

2 个答案: