RE: [PHP] Query Returning Error
| From: | Graham Cossey | Date: | Wed, 13 Oct 2004 10:48:03 +0000 |
| Subject: | RE: [PHP] Query Returning Error | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-199351@lists.php.net to get a copy of this message | ||
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
> -----------------------------
>
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