From: | Wes James <comptekki(at)gmail(dot)com> |
---|---|
To: | Justin Graf <justin(at)magwerks(dot)com> |
Cc: | pgsql-sql(at)postgresql(dot)org |
Subject: | Re: how to construct sql |
Date: | 2010-06-02 19:52:54 |
Message-ID: | AANLkTimpIxZCKcSRRzNukug6hrYL8LrfYyKuWWcz0b3T@mail.gmail.com |
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Lists: | pgsql-sql |
On Wed, Jun 2, 2010 at 2:44 PM, Justin Graf <justin(at)magwerks(dot)com> wrote:
> On 6/2/2010 12:31 PM, Wes James wrote:
>> On Wed, Jun 2, 2010 at 10:55 AM, Oliveiros
>> <oliveiros(dot)cristina(at)marktest(dot)pt> wrote:
>>
>>> Hi,
>>> Have you already tried this out?
>>>
>>> select MAX(page_count_count) - MIN(page_count_count) from page_count group
>>> by page_count_pdate.
>>>
>>>
>>> Best,
>>> Oliveiros
>>>
>> Oliveiros,
>>
>> Thx that mostly works. I just tried it and on the days there is only
>> 1 entry it is 0 since max is the same as min so max - min is 0. Is
>> there a way to take in to account the 1 entry days?
>>
>> Again thx - I appreciate your help :)
>>
>> -wes
>>
>>
> Put in a case
>
> select
> case when MAX(page_count_count) - MIN(page_count_count)> 0 then
> MAX(page_count_count) - MIN(page_count_count)
> else
> MAX(page_count_count)
> from page_count
> group by page_count_pdate.
>
Thx it is closer (with an end in the case):
select
case when MAX(page_count_count) - MIN(page_count_count) > 0 then
MAX(page_count_count) - MIN(page_count_count)
else
MAX(page_count_count)
end as day_max
from page_count
group by page_count_pdate order by page_count_pdate;
the else puts out the total count on that day. I would need
max(page_count_count) - max(page_count_count_of_previous_day)
thx,
-wes
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