January 31, 2014 at 10:44 am
Hello -
As the subject states, I can't help but think there is a more efficient way to rewrite this query so that I don't have to perform the Counts twice, though my bag of SQL tricks is quite limited.
I am essentially trying to create a daily report which will grab the order counts for specific statuses for each user that has at least one order matching one of those statuses.
Any help is greatly appreciated!
-- Daily Report
SELECT u.UserID,
(SELECT COUNT(OrderID) FROM Orders WHERE OrderStatusID = 2) As Status2Count,
(SELECT COUNT(OrderID) FROM Orders WHERE OrderStatusID = 3) As Status3Count,
(SELECT COUNT(OrderID) FROM Orders WHERE OrderStatusID = 4) As Status4Count,
(SELECT COUNT(OrderID) FROM Orders WHERE OrderStatusID = 8) As Status8Count
FROM Users u
JOIN Orders o ON u.UserID = o.userID
WHERE (
((SELECT COUNT(*) FROM Orders WHERE OrderStatusID = 2) > 0) OR
((SELECT COUNT(*) FROM Orders WHERE OrderStatusID = 3) > 0) OR
((SELECT COUNT(*) FROM Orders WHERE OrderStatusID = 4) > 0) OR
((SELECT COUNT(*) FROM Orders WHERE OrderStatusID = 8) > 0)
)
GROUP BY u.UserID
January 31, 2014 at 11:09 am
There's an easier way and it's explained in this article: http://www.sqlservercentral.com/articles/T-SQL/63681/
This is an example that might not work because I have nothing to test on.
-- Daily Report
SELECT UserID,
SUM(CASE WHEN OrderStatusID = 2 THEN OrderCount END) As Status2Count,
SUM(CASE WHEN OrderStatusID = 3 THEN OrderCount END) As Status2Count,
SUM(CASE WHEN OrderStatusID = 4 THEN OrderCount END) As Status2Count,
SUM(CASE WHEN OrderStatusID = 8 THEN OrderCount END) As Status2Count,
FROM(
SELECT u.UserID,
o.OrderStatusID,
COUNT(*) OrderCount
FROM Users u
JOIN Orders o ON u.UserID = o.userID
WHERE OrderStatusID IN( 2, 3, 4, 8)
GROUP BY u.UserID, OrderStatusID
HAVING COUNT(*) > 0)x --This is a pre-aggregation to improve performance, but should be tested.
GROUP BY UserID
EDIT: I was evaluating wy word choice and easier might not be accurate when there's no practice on how to do it, but it's certainly more efficient and will come natural when you fully understand it.
January 31, 2014 at 11:52 am
Thank you so much Luis!
This worked perfectly, makes more sense, and I appreciate the link. 😀
January 31, 2014 at 12:17 pm
You're welcome. I hope you understand how it works. If you don't, please post any questions that you have.
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