Re: Baffled, line producing error

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 > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php >