MySQL INNER JOIN 从第二个表中只选择一行

您需要有一个子查询来获取每个 user ID 的最新日期。

SELECT  a.*, c.*
FROM users a 
    INNER JOIN payments c
        ON a.id = c.user_ID
    INNER JOIN
    (
        SELECT user_ID, MAX(date) maxDate
        FROM payments
        GROUP BY user_ID
    ) b ON c.user_ID = b.user_ID AND
            c.date = b.maxDate
WHERE a.package = 1

SELECT u.*, p.*
FROM users AS u
INNER JOIN payments AS p ON p.id = (
    SELECT id
    FROM payments AS p2
    WHERE p2.user_id = u.id
    ORDER BY date DESC
    LIMIT 1
)

或者

SELECT u.*, p.*
FROM users AS u
INNER JOIN payments AS p ON p.user_id = u.id
WHERE NOT EXISTS (
    SELECT 1
    FROM payments AS p2
    WHERE
        p2.user_id = p.user_id AND
        (p2.date > p.date OR (p2.date = p.date AND p2.id > p.id))
)

这些解决方案比 accepted answer 更好，因为它们在使用相同用户和日期进行多次付款时可以正常工作。您可以try on SQL Fiddle。

SELECT u.*, p.*, max(p.date)
FROM payments p
JOIN users u ON u.id=p.user_id AND u.package = 1
GROUP BY u.id
ORDER BY p.date DESC

看看这个sqlfiddle

   SELECT u.* 
        FROM users AS u
        INNER JOIN (
            SELECT p.*,
             @num := if(@id = user_id, @num + 1, 1) as row_number,
             @id := user_id as tmp
            FROM payments AS p,
                 (SELECT @num := 0) x,
                 (SELECT @id := 0) y
            ORDER BY p.user_id ASC, date DESC)
        ON (p.user_id = u.id) and (p.row_number=1)
        WHERE u.package = 1

你可以试试这个：

SELECT u.*, p.*
FROM users AS u LEFT JOIN (
    SELECT *, ROW_NUMBER() OVER(PARTITION BY userid ORDER BY [Date] DESC) AS RowNo
    FROM payments  
) AS p ON u.userid = p.userid AND p.RowNo=1

您的查询有两个问题：

每个表和子查询都需要一个名称，因此您必须将子查询命名为 INNER JOIN (SELECT ...) AS p ON ...。您拥有的子查询仅返回一个行期间，但实际上您希望每个行都有一个用户。为此，您需要一个查询来获取最大日期，然后自行加入以获取整行。

假设 payments.date 没有关系，请尝试：

    SELECT u.*, p.* 
    FROM (
        SELECT MAX(p.date) AS date, p.user_id 
        FROM payments AS p
        GROUP BY p.user_id
    ) AS latestP
    INNER JOIN users AS u ON latestP.user_id = u.id
    INNER JOIN payments AS p ON p.user_id = u.id AND p.date = latestP.date
    WHERE u.package = 1

@John Woo 的回答帮助我解决了类似的问题。我还通过设置正确的顺序改进了他的答案。这对我有用：

SELECT  a.*, c.*
FROM users a 
    INNER JOIN payments c
        ON a.id = c.user_ID
    INNER JOIN (
        SELECT user_ID, MAX(date) as maxDate FROM
        (
            SELECT user_ID, date
            FROM payments
            ORDER BY date DESC
        ) d
        GROUP BY user_ID
    ) b ON c.user_ID = b.user_ID AND
           c.date = b.maxDate
WHERE a.package = 1

不过，我不确定这有多有效。

SELECT U.*, V.* FROM users AS U 
INNER JOIN (SELECT *
FROM payments
WHERE id IN (
SELECT MAX(id)
FROM payments
GROUP BY user_id
)) AS V ON U.id = V.user_id

这将使它工作

Matei Mihai 给出了一个简单有效的解决方案，但只有在 SELECT 部分放置一个 MAX(date) 才会起作用，因此该查询将变为：

SELECT u.*, p.*, max(date)
FROM payments p
JOIN users u ON u.id=p.user_id AND u.package = 1
GROUP BY u.id

并且 order by 不会对分组产生任何影响，但它可以对 group by 提供的最终结果进行排序。我试过了，它对我有用。

如果您在 ORDER BY 子句中需要几个列，我的回答直接来自 @valex 非常有用。

    SELECT u.* 
    FROM users AS u
    INNER JOIN (
        SELECT p.*,
         @num := if(@id = user_id, @num + 1, 1) as row_number,
         @id := user_id as tmp
        FROM (SELECT * FROM payments ORDER BY p.user_id ASC, date DESC) AS p,
             (SELECT @num := 0) x,
             (SELECT @id := 0) y
        )
    ON (p.user_id = u.id) and (p.row_number=1)
    WHERE u.package = 1

这是非常简单的内部联接，然后按 user_id 分组，并在 payment_id 中使用 max 聚合函数，假设您的表是用户并且付款查询可以是

SELECT user.id, max(payment.id)
FROM user INNER JOIN payment ON (user.id = payment.user_id)
GROUP BY user.id