I'm not sure if it's right solution but I did this -
$ins_u = mysql_query("INSERT INTO users (name,email)
VALUES ('$u_name','$u_email')");
$userid = mysql_insert_id();
instead for this -
$ins_u = @mysql_query("INSERT INTO users (name,email)
VALUES ('$u_name','$u_email')");
$userid = @mysql_insert_id($ins_u);
And it works perfectly
Youri
On 24 Aug 2001, at 20:51, Richard Lynch wrote:
> Your mysql_connect is wrong.
>
> Stop using just @ to suppress errors, and start doing something useful with
> http://php.net/mysql_error like sending it to http://php.net/error_log or
> something.
>
> --
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> ----- Original Message -----
> From: Brack <[EMAIL PROTECTED]>
> Newsgroups: php.general
> To: <[EMAIL PROTECTED]>
> Sent: Friday, August 24, 2001 3:25 PM
> Subject: HELP!!!
>
>
> > I have a script with combination:
> > $ins_u = @mysql_query("INSERT INTO users (name,email)
> > VALUES ('$u_name','$u_email')");
> > $userid = @mysql_insert_id($ins_u);
> >
> > It's working fine on local server but now I put it on the web and it
> > sais:
> > Warning: Supplied argument is not a valid MySQL-Link resource in
> > /home/sites/site21/web/incr/submition.inc on line 30
> > (which is line above)
> > Youri
> >
>
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