is correct, then where is the linkage between the two
SELECTs, and why not do the whole job in one joined SELECT?
Please advise,
=dn
- Original Message -
From: "Dave Carrera" <[EMAIL PROTECTED]>
To: "'DL Neil'" <[EMAIL PROTECTED]>
Sent: 16 March 2002
On Sat, 16 Mar 2002 11:50:11 -
"DL Neil" <[EMAIL PROTECTED]> wrote:
> Hi Dave,
> SELECT * FROM tblNm;
> SELECT * FROM tblNm GROUP BY Catid1;
> SELECT Catid1, count(*) FROM tblNm GROUP BY Catid1;
>
> As I said 'follow your instincts' and take it one step at a time: Code
> the simplest quer
On Sat, 16 Mar 2002 10:46:38 -
"Dave Carrera" <[EMAIL PROTECTED]> wrote:
> Hi All
>
> I am trying to count how many product names in my db have the same
> category id and then show it ie:
>
> Catid 1 Product 1
> Catid 1 Product 2
> Catid 2 Product 3
> Catid 3 Product 4
> Catid 3 Product 5
;error:
> ".mysql_error());
> $cntcat = mysql_fetch_array($cntresult,1,catcnt) ;{
> for($n =0; $n < count($cntcat); $n++){
> $showitemcount = each($cntcat);
>
> Dave Carrera
> Php Developer
> http://davecarrera.freelancers.net
> http://www.davecarrera.com
>
>
> -
Hi Dave,
> I am trying to count how many product names in my db have the same
> category id and then show it ie:
>
> Catid 1 Product 1
> Catid 1 Product 2
> Catid 2 Product 3
> Catid 3 Product 4
> Catid 3 Product 5
>
> Result would be
>
> Catid1 has 2 products
> Catid2 has 1 products
> Catid3 has
select cat_id, count(prod_id) from some_table order by cat_id;
Best regards,
Andrey Hristov
- Original Message -
From: "Dave Carrera" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Saturday, March 16, 2002 12:46 PM
Subject: [PHP-DB] a Count() ?
> Hi All
&
Hi All
I am trying to count how many product names in my db have the same
category id and then show it ie:
Catid 1 Product 1
Catid 1 Product 2
Catid 2 Product 3
Catid 3 Product 4
Catid 3 Product 5
Result would be
Catid1 has 2 products
Catid2 has 1 products
Catid3 has 2 products
I think i
