SQL Server - show rows for missing dates

Question

This is my SP in sql server 2008:

SELECT      
    CONVERT(NVARCHAR(30), YMD, 101) AS [DATE],               
    SUM(ISNULL(Jobs, 0)) AS [Jobs],     
    SUM         


        

Answer 1

I agree with @bluefeet approach

in order to create a dynamic from-to calendar you can use a recursive cte like this one:

DECLARE @DateFrom AS DATE
DECLARE @DateTo as DATE 
SET @DateFrom='7/23/2014'
SET @DateTo ='7/29/2014'
;WITH r AS
(SELECT @DateFrom AS calendardate
UNION ALL
SELECT DATEADD(d,1,calendardate) FROM r 
WHERE DATEADD(d,1,calendardate)<=@DateTo
) 
SELECT * FROM r

Answer 2

@Jayvee's approach is great if you're using a fairly short span (up to about 100 on most configurations) but can break down if you use too many. If you find yourself in that situation, use a WHILE loop. And here's a full example in case the JOIN isn't clear. Note that you said you were in a stored procedure, my alternative won't work in a view but his will.

DECLARE @DateFrom DATE
DECLARE @DateTo DATE
SET @DateFrom = '7/23/2014'
SET  @DateTo = '7/29/2014'

--BEGIN Add this
DECLARE @X DATE
DECLARE @D TABLE (EachDay Date)
SET @X = @DateFrom
WHILE @X <= @DateTo BEGIN
    INSERT INTO @D VALUES (@X)
    SET @X = DATEADD(day, 1, @X)
END
SELECT * FROM @D
--END Add this

SELECT      
    CONVERT(NVARCHAR(30), YMD, 101) AS [DATE],               
    SUM(ISNULL(Jobs, 0)) AS [Jobs],     
    SUM(ISNULL(SELL_VALUE, 0)) AS [COST]
    FROM    S       
        --ADD This - just the one following line
        RIGHT OUTER JOIN @D as D ON D.EachDay = S.YMD
    WHERE [YMD] BETWEEN @DateFrom AND @DateTo 
GROUP BY YMD
ORDER BY YMD DESC