Question

我有两个表，一个用于存储用户收入，另一个用于存储用户付款。我想做的是获得最低50美元的付款，并且只能从最后一次付款中获得。

表（userEarnings）包含以下列

id ->autoincrement  
userid -> unique id for users  
earn ->the amount of money the user earn (25.00, 14.50..)  
date_added -> DATETIME column

每次用户获得付款时，第二个表（userPayment）都会存储

pid->autoincrement  
uid ->unique user id, this is the same as userid from the above table  
paid_amt->payments the user received(500.00, 100.00...)  
paid_date ->the date the user was paid(DATETIME)

从这两张表中，如果金额超过50美元，我想列出自上次付款以来的用户ID和欠款总和。我假设我需要使用子查询和组，但不知道从哪里开始。

非常感谢任何帮助。感谢

Answer 1

试试这个：

SELECT userEarnings.userid, SUM(earn) AS total_earn FROM (
    SELECT uid, MAX(paid_date) AS paid_date
    FROM userPayment
    GROUP BY uid) AS T1
RIGHT JOIN userEarnings ON T1.uid = userEarnings.userid
WHERE date_added > T1.paid_date OR T1.paid_date IS NULL
GROUP BY userEarnings.userid
HAVING total_earn > 50

Answer 2

SELECT userEarnings.userid, SUM(userEarnings.earn) AS earn
FROM userEarnings 
LEFT OUTER JOIN (
    SELECT uid, MAX(paid_date) AS last_payment
    FROM userPayment
    GROUP BY uid) AS q
ON q.uid = userEarnings.userid
--the q.last_payment IS NULL clause ensures that the user who never got a payment before will get his first $50 check
WHERE (userEarnings.date_added > q.last_payment) OR (q.last_payment IS NULL)
GROUP BY userEarnings.userid
HAVING SUM(userEarnings.earn) >= 50.0

Answer 3

使用not exists子查询稍有不同的方法。可以以不同的方式执行，并且可能更容易阅读，具体取决于品味。

SELECT    ue.userid User
,         SUM(ue.earn) AmountDue
FROM      userEarnings ue
WHERE     NOT EXISTS (
          SELECT *
          FROM userPayment up
          WHERE up.uid = ue.userid
          AND up.paid_date >= ue.date_added
          )
GROUP BY  ue.userid
HAVING    AmountDue > 50.0

子查询以获得到期付款

3 个答案: