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MySQL - 基于同一个表中的行的总和列值

[英]MySQL - sum column value(s) based on row from the same table

我试图根据同一个表格中的ProductID在新列中获取“现金”,“检查”和“信用卡”总计。

表 - 付款

+-----------+------------+---------------+--------+
| ProductID |  SaleDate  | PaymentMethod | Amount |
+-----------+------------+---------------+--------+
|         3 | 2012-02-10 | Cash          |     10 |
|         3 | 2012-02-10 | Cash          |     10 |
|         3 | 2012-02-10 | Check         |     15 |
|         3 | 2012-02-10 | Credit Card   |     25 |
|         4 | 2012-02-10 | Cash          |      5 |
|         4 | 2012-02-10 | Check         |      6 |
|         4 | 2012-02-10 | Credit Card   |      7 |
+-----------+------------+---------------+--------+

期望的输出 -

+------------+------+-------+-------------+-------+
| ProductID  | Cash | Check | Credit Card | Total |
+------------+------+-------+-------------+-------+
|          3 |   20 |    15 |          25 |    60 |
|          4 |    5 |     6 |           7 |    18 |
+------------+------+-------+-------------+-------+

我试过LEFT JOINing同一张桌子,但没有取得任何成功。 任何建议,将不胜感激。 谢谢。

不成功和不完整的尝试 -

SELECT P.ProductID, Sum( PCash.Amount ) AS 'Cash', SUM( PCheck.Amount ) AS 'Check', SUM( PCredit.Amount) AS 'Credit Card' 
FROM Payments AS P 
LEFT JOIN Payments AS PCash ON P.ProductID = PCash.ProductID AND PCash.PaymentMethod = 'Cash'
LEFT JOIN Payments AS PCheck ON P.ProductID = PCheck.ProductID AND PCheck.PaymentMethod = 'Check'
LEFT JOIN Payments AS PCredit ON P.ProductID = PCredit.ProductID AND PCredit.PaymentMethod = 'Credit'
WHERE P.SaleDate = '2012-02-10' GROUP BY ProductID;

我认为你让它变得比它需要的复杂一点。

SELECT
    ProductID,
    SUM(IF(PaymentMethod = 'Cash', Amount, 0)) AS 'Cash',
    -- snip
    SUM(Amount) AS Total
FROM
    Payments
WHERE
    SaleDate = '2012-02-10'
GROUP BY
    ProductID

这可能被视为有点复杂但完全符合您的要求

SELECT 
  DISTINCT(p.`ProductID`) AS ProductID,
  SUM(pl.CashAmount) AS Cash,
  SUM(pr.CashAmount) AS `Check`,
  SUM(px.CashAmount) AS `Credit Card`,
  SUM(pl.CashAmount) + SUM(pr.CashAmount) +SUM(px.CashAmount) AS Amount
FROM
  `payments` AS p 
  LEFT JOIN (SELECT ProductID,PaymentMethod , IFNULL(Amount,0) AS CashAmount FROM payments WHERE PaymentMethod = 'Cash' GROUP BY ProductID , PaymentMethod ) AS pl 
    ON pl.`PaymentMethod` = p.`PaymentMethod` AND pl.ProductID = p.`ProductID`
  LEFT JOIN (SELECT ProductID,PaymentMethod , IFNULL(Amount,0) AS CashAmount FROM payments WHERE PaymentMethod = 'Check' GROUP BY ProductID , PaymentMethod) AS pr 
    ON pr.`PaymentMethod` = p.`PaymentMethod` AND pr.ProductID = p.`ProductID`
  LEFT JOIN (SELECT ProductID, PaymentMethod , IFNULL(Amount,0) AS CashAmount FROM payments WHERE PaymentMethod = 'Credit Card' GROUP BY ProductID , PaymentMethod) AS px 
    ON px.`PaymentMethod` = p.`PaymentMethod` AND px.ProductID = p.`ProductID`
GROUP BY p.`ProductID` ;

产量

ProductID | Cash | Check | Credit Card | Amount
-----------------------------------------------
    3     | 20   |  15   |   25        |  60
    4     | 5    |  6    |   7         |  18

SQL小提琴演示

SUM CASE使用示例:

    SELECT 
  DISTINCT(p.`ProductID`) AS ProductID,
  SUM(IF(p.`PaymentMethod`='Cash',Amount,0)) AS Cash_,
  SUM(IF(p.`PaymentMethod`='Check',Amount,0)) AS Check_,
  SUM(IF(p.`PaymentMethod`='Credit Card',Amount,0)) AS Credit_Card_,
  SUM( CASE PaymentMethod 
      WHEN 'Cash' THEN Amount
      WHEN 'Check' THEN Amount
      WHEN 'Credit Card' THEN Amount
     END) AS Total
FROM
  `payments` AS p 
GROUP BY p.`ProductID`;

SQL FIDDLE: http ://www.sqlfiddle.com/#!9 / 23d07d / 18

结果和表视图

暂无
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