Question

我有一个users表和一个payments表，对于每个有付款的用户，可能在payments表中有多个关联付款。我想选择所有付款用户，但只选择最新付款。我正在尝试这个SQL，但我从来没有尝试过嵌套的SQL语句，所以我想知道我做错了什么。感谢帮助

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

Answer 1

您需要有一个子查询才能获得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

Answer 2

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

查看此sqlfiddle

Answer 3

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
)

此解决方案比accepted answer更好，因为当某些付款使用相同的用户和日期时，它可以正常工作。

Answer 4

   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

Answer 5

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

这将使其正常工作

Answer 6

您的查询存在两个问题：

每个表和子查询都需要一个名称，因此您必须为子查询命名INNER JOIN (SELECT ...) AS p ON ...。
您拥有的子查询只返回一个行周期，但实际上您需要为每个用户创建一行。为此，您需要一个查询来获取最大日期，然后自我加入以获得整行。

假设payments.date没有联系，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

Answer 7

Matei Mihai给出了一个简单而有效的解决方案但是在SELECT部分放置MAX(date)之后它将无效，因此该查询将成为：

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

顺序不会对分组产生任何影响，但它可以命令分组提供的最终结果。我试过了，它对我有用。

Answer 8

@John Woo的回答帮助我解决了类似的问题。通过设置正确的顺序，我改进了他的答案。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

我不确定这是多么有效率。

Answer 9

如果在ORDER BY子句中需要多个cols，我的答案直接受到@valex的启发非常有用。

    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

Answer 10

您可以尝试以下方法：

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

Answer 11

这很简单：假设您的表是用户，并且付款查询可以是

，那么先进行内部联接，然后按user_id分组，并在payment_id中使用max汇总函数。

select user.id, max(payment.id) from user inner join payment on (user.id = payment.user_id) group by user.id

MySQL INNER JOIN只从第二个表中选择一行

11 个答案: