mysql - 从两个不同的行中为单个名称选择两个值

Question

我想从表中获取数据，其中有两个以上(至少 2 个是必须的)值是针对单个客户的，1) 债务 2) 已付债务。我需要根据同一行中每个客户的最大日期显示这两个值。我正在使用下面的查询

查询

SELECT MAX(RECEIPTS.DATENEW) AS DATE,
   MAX(CASE WHEN PAYMENTS.PAYMENT = 'debt' THEN PAYMENTS.TOTAL END) `TAKEN DEBT`,
   MAX(CASE WHEN PAYMENTS.PAYMENT = 'debtpaid' THEN PAYMENTS.TOTAL END) `PAID DEBT`,
   CUSTOMERS.NAME AS CUSTOMER
FROM RECEIPTS
INNER JOIN TICKETS ON RECEIPTS.ID = TICKETS.ID
    INNER JOIN PAYMENTS ON RECEIPTS.ID = PAYMENTS.RECEIPT
    INNER JOIN CUSTOMERS ON TICKETS.CUSTOMER = CUSTOMERS.ID
WHERE PAYMENTS.PAYMENT IN('debt', 'debtpaid')
GROUP BY RECEIPTS.ID

返回结果如下，看起来更接近...

但我需要将单个客户的债务和已偿还债务值放在同一行中，并且还需要他们的日期(个人日期)。所需数据如下图所示

Answer 1

好吧，不知怎么的，这奏效了! 需要注意几件事，正如我所说，每个客户都有两个条目，因此必须按 CUSTOMER 分组并对日期应用相同的 CASE 子句，因此只需放置代码以防有人发现它有帮助。

SELECT
    MAX(CASE WHEN PAYMENTS.PAYMENT = 'debt' THEN PAYMENTS.TOTAL END) AS TAKENDEBT,
    MIN(CASE WHEN PAYMENTS.PAYMENT = 'debtpaid' THEN PAYMENTS.TOTAL END) AS GIVENDEBT,
    MAX(CASE WHEN PAYMENTS.PAYMENT = 'debt' THEN RECEIPTS.DATENEW END) AS TAKENDATE,
    MAX(CASE WHEN PAYMENTS.PAYMENT = 'debtpaid' THEN RECEIPTS.DATENEW END) AS GIVENDATE,
    CASE WHEN PAYMENTS.PAYMENT = 'debt' OR PAYMENTS.PAYMENT = 'debtpaid' THEN CUSTOMERS.NAME END AS CUSTOMER 
FROM RECEIPTS
    INNER JOIN TICKETS ON RECEIPTS.ID = TICKETS.ID
    INNER JOIN PAYMENTS ON RECEIPTS.ID = PAYMENTS.RECEIPT
    INNER JOIN CUSTOMERS ON TICKETS.CUSTOMER = CUSTOMERS.ID
WHERE
    (PAYMENTS.PAYMENT = 'debt'
    OR PAYMENTS.PAYMENT = 'debtpaid')
GROUP BY CUSTOMER
ORDER BY CUSTOMER

感谢您的所有建议!