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SQL 2008 Stored Procedure - multiple counts

Posted on 2011-02-15
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Last Modified: 2012-06-21
Appreciate some help;

Sql 2008 stored procedure or Select statement preferred.
Say I have a 2 column table. Field 1 = Office and Field 2 = DateDue
I wish to count all those records which are earlier than a date (for example lets say GetDate())
I need to group the counts on the Office field.
I need to have 3 separate counts for each office. Less then now, less then now +1 month, and less than now +2 months.

So the result would be a mini recordset

Office   Count1   Count2   Count3
1         24            20             0
2         12            23             2
3         0              10             34
4         34             8              0

Can this be done in a single select statement? Perhaps temp table SP perhaps?

Thx
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Question by:BrightRaven
2 Comments
 
LVL 11

Accepted Solution

by:
jasonduan earned 500 total points
ID: 34899184
SELECT Office,
         SUM(C1) AS Count1,
         SUM(C2) AS Count2,
         SUM(C3) AS Count3
FROM
(
    SELECT Office,
         CASE WHEN DateDue <= GetDate() THEN 1 ELSE 0 END AS C1,
         CASE WHEN DateDue <= DateAdd(month, 1, GetDate()) THEN 1 ELSE 0 END AS C2,
         CASE WHEN DateDue <= DateAdd(month, 2, GetDate()) THEN 1 ELSE 0 END AS C3,
    FROM MyTable              
) T
GROUP BY Office
0
 
LVL 50

Expert Comment

by:Lowfatspread
ID: 34899563

select office,[0] as now,[1] as next month,[2] as [2months]
 from (
select office,case when duedate < convert(char(8),getdate(),112)+' 23:59:59.997' then 0
                        when duedate < dateadd(m,+1,convert(char(8),getdate(),112)+' 23:59:59.997' then 1
     else 2 end as Due
 from yourtable
 
 where duedate <= dateadd(m,1,convert(char(8),getdate(),112)+' 23:59:59.997')
) as x
pivot (count(*) for due in ([0],[1],[2])) as pvt
order by office
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