Question

PaymentHistory表

==========================
UserId | Amount | Date
==========================
 1     | 300    | 2014-06-26
 3     | 300    | 2014-06-26
 2     | 200    | 2014-06-26
 1     | 400    | 2014-06-22
 1     | 100    | 2014-06-21

首先显示两列是金额之和作为今天具有相同日期的命名“currentEarning”，第二个是基于今天的前一日期命名为“oldEarning”的金额之和。（对于特定的userId）请参阅输出：

输出

=====================================
UserId | currentEarning | oldEarning |
=====================================
 1     | 300            | 500

Answer 1

您可以使用一个查询执行此操作：

SELECT
    SUM(IF(`Date` = CURRENT_DATE, AMOUNT, 0)) AS currentEarning,
    SUM(IF(`Date` < CURRENT_DATE, AMOUNT, 0)) AS oldEarning
FROM PaymentHistory
GROUP BY `UserId`

比子查询或两个单独的查询更有效。

Answer 2

获取当前金额：

SELECT UserId, SUM(Amount) as currentEarning, Date FROM PaymentHistory WHERE UserId = '1' AND DATE(Date) = CURDATE()

获取旧金额：

SELECT UserId, SUM(Amount) as oldEarning, Date FROM PaymentHistory WHERE UserId = '1' AND DATE(Date) != CURDATE()

这应该有效

Answer 3

SELECT `UserId`, 
    (SELECT SUM(`Amount`) FROM `PaymentHistory` WHERE `UserID` = t1.`UserId` AND `Date` = t1.`Date`) AS `currentEarning`,
    (SELECT SUM(`Amount`) FROM `PaymentHistory` WHERE `UserID` = t1.`UserId` AND `Date` < t1.`Date`) AS `oldEarning`
FROM PaymentHistory AS t1
WHERE t1.`Date` = CURRENT_DATE
GROUP BY t1.`UserId`
HAVING `oldEarning` IS NOT NULL

根据日期获取总金额

3 个答案: