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

Solution 1:

I think you're making this a bit more complicated than it needs to be.

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

Solution 2:

This might be seen as a little complex but does exactly what you want

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` ;

Output

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

SQL Fiddle Demo

Solution 3:

SUM CASE using example:

    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