I have this query, it gives the suspected information in myphpadmin:

select listid, count(*) from table1 group by listid

How do I put this the result into an array in php? I have problems with the
count(*) field

Jaak wrote:
> I have this query, it gives the suspected information in myphpadmin:
>
> select listid, count(*) from table1 group by listid
>
> How do I put this the result into an array in php? I have problems with the
> count(*) field
>
>

$q='select listid, count(*) from table1 group by listid';
$res=mysql_query($q);
if($res){
$data=array();
while($row=mysql_fetch_array($res,MYSQL_NUM)){
$data[]=array(
'listid' => $row[0],
'count' => $row[1]
);
}
}

Untested, but it should give you the idea... Another option is to change
the query to something similar to:

select listid, count(*) as num_found from table1 group by listid

Then if you are using mysql_fetch_assoc, then the key name will be
'num_found'

HTH

--
Justin Koivisto, ZCE - justin@koivi.com
http://koivi.com

"Justin Koivisto" <justin@koivi.com> schreef in bericht
news:dLGdnV0pDpb1jFjeRVn-rg@onvoy.com...
> Jaak wrote:
>> I have this query, it gives the suspected information in myphpadmin:
>>
>> select listid, count(*) from table1 group by listid
>>
>> How do I put this the result into an array in php? I have problems with
>> the
>> count(*) field
>>
>>
>
> $q='select listid, count(*) from table1 group by listid';
> $res=mysql_query($q);
> if($res){
> $data=array();
> while($row=mysql_fetch_array($res,MYSQL_NUM)){
> $data[]=array(
> 'listid' => $row[0],
> 'count' => $row[1]
> );
> }
> }
>
> Untested, but it should give you the idea... Another option is to change
> the query to something similar to:
>
> select listid, count(*) as num_found from table1 group by listid
>
> Then if you are using mysql_fetch_assoc, then the key name will be
> 'num_found'
Thanks Justin, it really helped me! I was confused with the 'count' thing,
because this was not really a field in the table.