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How can I return a count of one when there are more than one in the count?

Select Order#, Date

   Order#                    Date
         3456                01/15/2013
         3456                01/18/2013
         3456                01/30/2013
         3457                01/25/2013
         3458                02/15/2013

Select Order#, Count(date)
group by Order#

   Order#                    Date
         3456                        3        
         3457                        1
         3458                        1

Anytime there is more than one date, as in 3456 above, I need to only return a count of 1.
How can I do this?
0
rhservan
Asked:
rhservan
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1 Solution
 
pjevinCommented:
Select Order#, 1 as Date group by Order#
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Aneesh RetnakaranDatabase AdministratorCommented:
Select Order#, 1 as date
group by Order#
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Jim HornMicrosoft SQL Server Developer, Architect, and AuthorCommented:
<potentially stupid question>

>Anytime there is more than one date, as in 3456 above, I need to only return a count of 1.
Then what's the purpose of having a count?
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rhservanAuthor Commented:
Simply when the query returns more than 1 date then I need the other dates ignored.
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Jim HornMicrosoft SQL Server Developer, Architect, and AuthorCommented:
Then it looks like hard-coding 1 would be the correct answer, as the first two experts posted.

Or, you don't even need the 1..

SELECT DISTINCT  Order#  FROM YourTable

Open in new window

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rhservanAuthor Commented:
But I still need the date value returned.
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Jim HornMicrosoft SQL Server Developer, Architect, and AuthorCommented:
>But I still need the date value returned.
Okay, but I think you need to eyeball the desired return recordset in your question, and make sure it's correct, as I don't see any dates in it.  

For example, for 3456, what would the date value be you want returned:  01/15/2013 (min), 01/18/2013, or 01/30/2013 (max)?
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pjevinCommented:
As in an actual date or your value of 1?

Select Order#, 1 as Date group by Order#  gives you your

Order#                    Date
         3456                        1        
         3457                        1
         3458                        1

Select Order#, Max(Date) as Date group by Order#  gives you one date value per order number (the max date value in this case).

Order#                    Date
         3456                        01/30/2013        
         3457                        01/25/2013
         3458                        02/15/2013
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PortletPaulfreelancerCommented:
>>what's the purpose of having a count?
to invent new ways of getting there?

select
  order#
, (count(*)+1) - count(*) as forced_to_one_the_hard_way_option_1

, case when count(*) > 0 then count(*) / count(*)  
            else  (count(*)+1) / (count(*)  +1)
            end                         as forced_to_one_the_hard_way_option_2

, 1                                         as forced_to_one_the_fixed_way
from yourtable
group by order#

-- sorry, but a count is a count, apologies for the flippancy in the above
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PortletPaulfreelancerCommented:
ok, flippancy aside now. In many questions what emerges eventually is that "the most recent" record is required (i.e. the full record matching the most recent date), so maybe this will help?
select
  OrderID
, row_ref
from (
       select
          *
       , row_number() over (partition by OrderID order by OrdDate DESC) as row_ref
       from orders
     ) as o
where row_ref = 1
order by 
         OrdDate

Open in new window

it produces:
ORDERID      ROW_REF
3457      1
3456      1
3458      1

see it at: http://sqlfiddle.com/#!3/2adec/2

& :) with flippancies at work: http://sqlfiddle.com/#!3/2adec/3 (couldn't help it)
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Jim HornMicrosoft SQL Server Developer, Architect, and AuthorCommented:
My compliments on an excellent flippancy.
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