asked on

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?

pjevin

Select Order#, 1 as Date group by Order#

Aneesh

Select Order#, 1 as date
group by Order#

Jim Horn

<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?

rhservan

ASKER

Simply when the query returns more than 1 date then I need the other dates ignored.

Jim Horn

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

rhservan

ASKER

But I still need the date value returned.

Jim Horn

>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)?

ASKER CERTIFIED SOLUTION

pjevin

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PortletPaul

>>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

PortletPaul

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)

Jim Horn

My compliments on an excellent flippancy.