Re: $result = mysql_query($query);
| From: | Daniel Gosselin | Date: | Sat, 14 Sep 2002 19:42:55 +0000 |
| Subject: | Re: $result = mysql_query($query); | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-116397@lists.php.net to get a copy of this message | ||
I read on php.net that you have to use SELECT count(*)
with if(mysql_num_rows($result) != 0) but if i use count(*)
and i have no result to show the tile is printed with empty result
so i'll use if($row = mysql_fetch_array($result)), work fine,
----- Original Message -----
From: "John Holmes" <holmes072000@charter.net>
To: "'DtM'" <info@protento.com>; <php-general@lists.php.net>
Sent: Saturday, September 14, 2002 3:05 PM
Subject: RE: [PHP] $result = mysql_query($query);
> > $query = "SELECT * from TABLE WHERE CHAMP = 'truc' order by NO LIMIT
> > 0,10";
> > $result = mysql_query($query);
> >
> > what i'm trying to do is if the result is !=0 echo the result else
> echo
> > nothing
> > i tried this
>
> OK, let's think about this. You want to show a result if rows are
> returned...
> So...
>
> if($row = mysql_fetch_array($result))
> {
> do
> {
> //do whatever
> }while($row = mysql_fetch_array($result));
> }
> else
> { //no results }
>
> That'll work fine. Or use mysql_num_rows(), like others suggested.
>
> ---John Holmes...