Question

我有两个简单的SELECT查询 - 加入MySQL DB中的同一个表：

SELECT uID, externaluID, COUNT(editID) AS editCount 
FROM users
LEFT JOIN (edits
    INNER JOIN posts 
    ON postRefID = postID AND editAuthorID <> authorID AND isa = 0)
ON editAuthorID = uID
GROUP BY uID
ORDER BY uID;

SELECT uID, externaluID, COUNT(posts.postID) AS postCount, SUM(value)
FROM users
LEFT JOIN (posts
    LEFT JOIN usrR ON posts.postID = usrR.postID)
ON authorID = uID
GROUP BY authorID
ORDER BY uID;

到目前为止一切顺利。现在我想合并这些查询。我的方法是

SELECT uID, externaluID, COUNT(editID) AS editCount, COUNT(P2.postID) AS postCount, SUM(rateValue)
FROM users
LEFT JOIN (edits
    INNER JOIN posts AS P1
    ON postRefID = P1.postID AND editAuthorID <> P1.authorID AND isa = 0)
ON editAuthorID = uID
LEFT JOIN (posts AS P2
    LEFT JOIN usrR ON P2.postID = usrR.postID)
ON P2.authorID = uID
GROUP BY P2.authorID, uID
ORDER BY uID;

但它返回错误的结果。我做错了什么？

Answer 1

试试这个：

SELECT u1.uID, u1.externaluID, COUNT(e1.editID) AS editCount, u2.postCount, u2.sumValue
FROM users u1 
LEFT JOIN (edits INNER JOIN posts 
        ON postRefID = postID AND editAuthorID <> authorID AND isa = 0) as e1
    ON e1.editAuthorID = u1.uID
LEFT JOIN (
    SELECT uID, externaluID, COUNT(posts.postID) AS postCount, SUM(value) as sumValue
    FROM users
    LEFT JOIN (posts
        LEFT JOIN usrR ON posts.postID = usrR.postID)
    ON authorID = uID
    GROUP BY authorID
    ORDER BY uID) as u2 on u1.uID = u2.uID 
GROUP BY u1.uID,u2.postCount, u2.sumValue
ORDER BY u1.uID

合并两个包含JOIN的SELECT

1 个答案: