RE: [PHP] Query Returning Error

Thanks Chris, that'll help keep my code a bit more 'compact'. (Sorry Harlequin, kinda hijacked your post here) Graham > -----Original Message----- > From: Chris Dowell [mailto:php@nodalplane.com] > Sent: 13 October 2004 11:57 > To: php-general@lists.php.net > Subject: Re: [PHP] Query Returning Error > > > You can get around this problem more easily by putting your variables > inside curly brackets when they appear inside strings. > > See here: > > http://uk.php.net/manual/en/language.types.string.php#language.typ > es.string.parsing.complex > > Or as an example: > > <?php > > $Query01 = "SELECT * FROM Users > WHERE UserID = '{$_POST['TXT_UserID']}' > AND UserPassword = '{$_POST['TXT_UserPassword']}'"; > > $Result01 = mysql_query($Query01) or die ("Error 01: " . mysql_error()); > > ?> > > Note also that you didn't quote the index you were using inside the > $_POST array. This is wrong - it will work in "raw" PHP because it will > be interpreted as an undefined constant, and thus replaced with the > string, however the behaviour inside a quoted string is even less > correct, and will almost certainly not do what you expect. > > Hope this helps > > Cheers > > Chris > > > Graham Cossey wrote: > > Besides adding $ to Query01, I have had little luck in the past > when using > > $_POST or $_GET directly in an evaluated string. If you are still having > > problems, try something like: > > > > <?php > > $user = $_POST['TXT_UserID']; > > $pwd = $_POST['TXT_UserPassword']; > > > > // Authenticate User: > > $Query01 = "SELECT * FROM Users > > WHERE UserID='$user' AND > > UserPassword='$pwd'"; > > $Result01 = mysql_query($Query01) or die("Error 01: " . > mysql_error()); > > ?> > > > > HTH > > > > Graham > > > > > >>-----Original Message----- > >>From: Harlequin [mailto:michael.mason@arraspeople.co.uk] > >>Sent: 13 October 2004 08:32 > >>To: php-general@lists.php.net > >>Subject: [PHP] Query Returning Error > >> > >> > >>Morning all. > >> > >>this is such a basic question I'm embarrassed to ask but the > query worked > >>fine a few minutes ago and now returns an error: > >> > >>I get an error: > >> > >>Parse error: parse error, unexpected '=' in sample.php on line 2 > >> > >>[CODE] > >>// Authenticate User: > >> Query01 = "SELECT * FROM Users > >> WHERE UserID='$_POST[TXT_UserID]' > >> AND UserPassword='$_POST[TXT_UserPassword]'"; > >> $Result01 = mysql_query($Query01) or die("Error 01: " . > mysql_error()); > >>[/CODE] > >> > >>WTF...? > >> > >>-- > >>----------------------------- > >> Michael Mason > >> Arras People > >> www.arraspeople.co.uk > >>----------------------------- > >> > >>-- > >>PHP General Mailing List (http://www.php.net/) > >>To unsubscribe, visit: http://www.php.net/unsub.php > >> > >> > > > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > >