SQL group by day, show orders for each day

Question

I have an SQL 2005 table, let\'s call it Orders, in the format:

OrderID, OrderDate,  OrderAmount
1,       25/11/2008, 10
2,       25/11/2008, 2
3,       30/1         


        

Answer 1

SQL isn't "skipping" dates... because queries run against data that is actually in the table. So, if you don't have a DATE in the table for January 14th, then why would SQL show you a result :)

What you need to do is make a temp table, and JOIN to it.

CREATE TABLE #MyDates ( TargetDate DATETIME )
INSERT INTO #MyDates VALUES CONVERT(DATETIME, CONVERT(VARCHAR, GETDATE() - 0, 101))
INSERT INTO #MyDates VALUES CONVERT(DATETIME, CONVERT(VARCHAR, GETDATE() - 1, 101))
INSERT INTO #MyDates VALUES CONVERT(DATETIME, CONVERT(VARCHAR, GETDATE() - 2, 101))
INSERT INTO #MyDates VALUES CONVERT(DATETIME, CONVERT(VARCHAR, GETDATE() - 3, 101))
INSERT INTO #MyDates VALUES CONVERT(DATETIME, CONVERT(VARCHAR, GETDATE() - 4, 101))
INSERT INTO #MyDates VALUES CONVERT(DATETIME, CONVERT(VARCHAR, GETDATE() - 5, 101))
INSERT INTO #MyDates VALUES CONVERT(DATETIME, CONVERT(VARCHAR, GETDATE() - 6, 101))
INSERT INTO #MyDates VALUES CONVERT(DATETIME, CONVERT(VARCHAR, GETDATE() - 7, 101))

SELECT CONVERT(VARCHAR, TargetDate, 101) AS Date, COUNT(*) AS OrderCount
FROM dbo.Orders INNER JOIN #MyDates ON Orders.Date = #MyDates.TargetDate
GROUP BY blah blah blah (you know the rest)

There you go!