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Posted on 2012-03-18
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Last Modified: 2012-03-28
Hello,

I have a table called Documents that holds a number of documents.  There are couple of columns that are interesting to me.  

The first is doc_Datetime that holds the datetime that the document was added.

The second is the doc_Exported (bit) that tells me the document has been exported.

I need help with a query that will show the total number of documents recieved each day, and the number of them that have been exported.

The result set should even include days when no records have been received.

Something like:

Date                  Documents received   Documents exported.
2012-03-18       123                               80
2012-03-19       10                                 9
2012-03-20       0                                   0
2012-03-21       12                               11

Thanks!
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Question by:soozh
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2 Comments
 
LVL 42

Expert Comment

by:dqmq
ID: 37735181
the basic query is easy:

select date, count(*), sum(exported) as exported
from
(select  convert (varchar(10), doc_datetime ,101) as Date, case when doc_exported = 1 then 1 else 0 end as exported from YOURTABLE) as tmp
group by date
order by date

Returning rows for missing dates is tricky, I'll need to get back to you about that.  Do you have a table of dates, perchance?
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LVL 42

Accepted Solution

by:
dqmq earned 2000 total points
ID: 37735224
try this:

;with cte as
(select min(cast(convert (varchar(10), doc_datetime ,101) as datetime)) as date
      , max(cast(convert (varchar(10), doc_datetime ,101) as datetime)) as max
 from YOURTABLE
 union all
 select date + 1, max from cte where date < max)

select cte.date, count(tmp.date) as received, isnull(sum(exported),0) as exported
from
(select cast(convert (varchar(10), doc_datetime ,101) as datetime) as Date, case when doc_exported = 1 then 1 else 0 end as exported from YOURTABLE) as tmp
right join cte on tmp.date = cte.date
group by cte.date
order by cte.date
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