Do not use double kotes here.
$link_glob = \"mysql_connect(\'$host_glob\', \'$un_glob\',
\'$pw_glob\')\";
mysql_connect is a function.
Regards,
Laercio Fortes
Citando Jason Soza <[EMAIL PROTECTED]>:
> If this is truly the code you\'re using, you\'re missing the closing
> curly-brace after the else statement. I.e. this:
> } else {
> echo \"<font size=+1>Your Information has successfully been
> entered into the database!</font><br>\";
>
> Should be this:
> } else {
> echo \"<font size=+1>Your Information has successfully been
> entered into the database!</font><br>\";
> }
>
> HTH,
> Jason Soza
>
> ----- Original Message -----
> From: Jule Slootbeek <[EMAIL PROTECTED]>
> Date: Monday, June 3, 2002 2:09 pm
> Subject: stupid error, please kick me (and send me a solution)
>
> > Hey guys,
> > i\'m getting this error with the following sql script using php:
> > --error--
> > Warning: Supplied argument is not a valid MySQL-Link resource in
> > /var/www/phpquiz/register_user.php on line 12
> > --error--
> >
> > --script--
> > $link_glob = \"mysql_connect(\'$host_glob\', \'$un_glob\', \'$pw_glob\')\";
> > $query = \"INSERT INTO user values(\'0\', \'$fname\', \'$lname\',
> > \'$email\',
> > \'$username\', PASSWORD(\'$password\')\";
> > $result = mysql_db_query(\'$db_glob\', \'$query\',
> > $link_glob\'); if (!$result) {
> > echo \"<font size=+1>Your Information
> > could not be entered into the database,
> > Please contact the
> > href=mailto:$webmaster>webmaster.</font><br><br>\" . mysql_errno()
> > .
> > \": \" . mysql_error() . \"<br><br>\";
> > } else {
> > echo \"<font size=+1>Your Information
> has
> > successfully been entered into the
> > database!</font><br>\";
> >
> > --
> > Jule Slootbeek
> > [EMAIL PROTECTED]
> >
> > http://blindtheory.cjb.net
>
>
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