You have:
if ($player_password != $player_password_verify) {
$errmsg .= 'Password don't match. Please try again<BR>';
$error = 1;
}
Notice the ' inside the '', this is bad syntax. For more
information on using strings in PHP, see:
http://www.zend.com/zend/tut/using-strings.php
http://www.php.net/manual/en/language.types.string.php
One thing you can do is escape it: \'
regards,
Philip Olson
On Sun, 2 Jun 2002, Craig Vincent wrote:
> > If the "error" is a warning about undefined variable, then set a default
> > value for $errmsg before you start adding strings to it.
> >
> > $errmsg .= "this";
> >
> > That by itself means $errmsg = $errmsg . "this";, but if $errmsg isnt'
> > defined, you'll get the warning.
> >
> > Set $errmsg = ''; at the beginning of your script if that is the
> > problem...
> >
> > For future reference, always give the exact error when posting.
> >
> > Trying to be psychic,
>
> You'll notice a few lines up I have defined $errmsg =) It's a standard
> parsing error I'm getting
>
> Parse error: parse error in admin_add_player.php on line 15
>
> Since this message does not arise when line 15 is removed I can only assume
> the error is actually on that line and not a missing quote or bracket
> somewhere else in the script.
>
> Sincerely,
>
> Craig Vincent
>
>
>
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