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

I had the same problem and this is how I solved it:

SELECT datename(DW,nDays) TimelineDays, 
    Convert(varchar(10), nDays, 101) TimelineDate,
    ISNULL(SUM(Counter),0) Totals 
FROM (Select GETDATE() AS nDays
    union Select GETDATE()-1
    union Select GETDATE()-2
    union Select GETDATE()-3
    union Select GETDATE()-4
    union Select GETDATE()-5
    union Select GETDATE()-6) AS tDays

Left Join (Select * From tHistory Where Account = 1000) AS History
            on (DATEPART(year,nDays) + DATEPART(MONTH,nDays) + DATEPART(day,nDays)) = 
            (DATEPART(year,RecordDate) + DATEPART(MONTH,RecordDate) + DATEPART(day,RecordDate)) 
GROUP BY nDays
ORDER BY nDays DESC

The ouput is:

TimelineDays,   TimelineDate,     Totals

Tuesday         10/26/2010        0
Monday          10/25/2010        6
Sunday          10/24/2010        3
Saturday        10/23/2010        2
Friday          10/22/2010        0
Thursday        10/21/2010        0
Wednesday       10/20/2010        0