Tell me more ×
Stack Overflow is a question and answer site for professional and enthusiast programmers. It's 100% free, no registration required.

I'm trying to get 'Cash', 'Check' and 'Credit Card' totals in new columns based on ProductID from the same table.

Table - Payments

+-----------+------------+---------------+--------+
| 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 |
+-----------+------------+---------------+--------+

Desired Output -

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

I've tried LEFT JOINing the same table but haven't had any success. Any suggestions would be appreciated. Thanks.

Unsuccessful and incomplete attempt -

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;
share|improve this question

2 Answers

up vote 6 down vote accepted

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
share|improve this answer
there are no check and c_card column in fetch result – diEcho Feb 11 at 4:43
1  
@diEcho I didn't type that part out because I assumed that he could figure that out on his own without any trouble – Explosion Pills Feb 11 at 4:45
Thank you so much. I knew there had to be an easier way. – newbie Feb 11 at 5:48

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

share|improve this answer
Thanks for the response. I see I was missing the 2nd select statement after LEFT JOIN. Would give you a vote up but 'Vote Up requires 15 reputation' and I'm 4 away! :) Thanks. – newbie Feb 11 at 17:49
Welcome! No problemo – raheel shan Feb 11 at 17:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.