RE: [PHP] $result = mysql_query($query);
| From: | Jeremy Wilson | Date: | Sat, 14 Sep 2002 23:52:31 +0000 |
| Subject: | RE: [PHP] $result = mysql_query($query); | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-116406@lists.php.net to get a copy of this message | ||
The correct way to do this would be the following;
$query = "SELECT * FROM TABLE WHERE THIS='THAT'";
$res = mysql_query($query);
If (mysql_num_rows($res) > 0) {
// there are results
}
Else {
// there are no results
}
-----Original Message-----
From: Daniel Gosselin [mailto:info@protento.com]
Sent: Saturday, September 14, 2002 12:43 PM
To: holmes072000@charter.net; php-general@lists.php.net
Subject: Re: [PHP] $result = mysql_query($query);
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...
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