Re: HELP!!!

From: Date: Sat, 25 Aug 2001 08:22:56 +0000
Subject: Re: HELP!!!
References: 1  Groups: php.general 
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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. > > -- > WARNING richard@zend.com address is an endangered species -- Use > ceo@l-i-e.com > Wanna help me out? Like Music? Buy a CD: > http://l-i-e.com/artists.htm > Volunteer a little time: http://chatmusic.com/volunteer.htm > ----- Original Message ----- > From: Brack <brak@nettaxi.com> > Newsgroups: php.general > To: <php-general@lists.php.net> > 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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