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:[EMAIL PROTECTED]
> Sent: 13 October 2004 11:57
> To: [EMAIL PROTECTED]
> 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:[EMAIL PROTECTED]
> >>Sent: 13 October 2004 08:32
> >>To: [EMAIL PROTECTED]
> >>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
> >>
> >>
> >
> >
>
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