Php 根据日期获取总金额

付款历史记录表

==========================
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 

显示两列，第一列是与今天日期相同的名为currentEarning的金额总和，第二列是基于今天以前日期的名为oldEarning的金额总和。有关特定用户ID，请参见输出：

输出

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

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()

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这应该可以用

只需一个查询即可：

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

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

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比子查询或两个单独查询更高效。

与子句一起使用。sql在哪里？你的php在哪里？@Fred ii-我知道这里使用的求和函数，但实际上我想根据各种条件生成两列，我需要生成一列query@Z好的，嗯，那么你应该使用内部连接和外部连接，但我不太擅长：抱歉，同意。出于某种原因，我的脑海中有这样一个想法，那就是，这个小组将无法正常工作。时间还早；