On 05-Feb-2001 Steve Werby wrote:
> <[EMAIL PROTECTED]> wrote:
>> i thought about that, but i gave up on my first attempt on doing so,
> because
>> mysql complained about my query, it seems that it doesn't allow to use the
>> max() function with a sum() function, an example of the query is:
>>
>> SELECT month,
>> daynr,
>> hour,
>> sum(numberofusers),
>> sum(numberofmails),
>> sum(numberofdownloads)
>> FROM table
>> GOURP BY month, daynr, hour
>> WHERE where_clause;
>>
>> and mysql complaints when i do another column with max(sum(numberofusers))
>:\
>
> You just need to add a column max(numberofusers). You've already grouped by
> month, day and hour so:
>
> sum(numberofusers) => total # of users for that month/day/hour/
> max(numberofusers) => max # of users for any record for that month/day/hour
>
> If you're really trying to get the total # of users for the single
> month/day/hour with the most users then you need a second query. If you can
> put into sentences exactly what you're trying to do, we can probably help
> you a little better.
>
>> i think i'm doomed with the big loop thing :\
>
> Nah, I think you can avoid it. MySQL is your friend. <grin>
>
> --
> Steve Werby
> COO
> 24-7 Computer Services, LLC
> Tel: 804.817.2470
> http://www.247computing.com/
>
>
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ok, this is the real query:
select sum(numberofusers),
sum(numberofmails),
sum(numberofdownloads),
month(data) as month,
hour
from account, site
where site.pid=account.id and site.siteid=10
and hour!=''
group by month,hour order by month,hour
while($db->next_record())
row[$db->Record[3]][$db->Record[4]]=$db->Record;
}
then display the table
| |month1 |month2 |
|hour1 |value |value |
|hour2 |value |value |
and to distinguish the maximum value and minimum value with somekind of color
or bold or something.
this is the whole job, i can't seem to find another way beside the loop
thing...
thanks a lot for the help guys.
Rui Barreiros
Software Developer
WEBSOLUT - Soluções Internet
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