[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
Re: [PHP] Query Returning Error
On Wednesday 13 October 2004 15:32, Harlequin wrote: 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: Change everything back to what it was a few minutes ago? -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-general -- /* Murphy was a grunt -- Murphy's Military Laws n56 */ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Query Returning Error
Doh...! thanks mate. -- - Michael Mason Arras People www.arraspeople.co.uk - Jason Wong [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED] On Wednesday 13 October 2004 15:32, Harlequin wrote: 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: Change everything back to what it was a few minutes ago? -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-general -- /* Murphy was a grunt -- Murphy's Military Laws n56 */ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Query Returning Error
should Query01 have a $ in front of it, i assume its a var Jason Harlequin [EMAIL PROTECTED] wrote: 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
Re: [PHP] Query Returning Error
On Wed, 13 Oct 2004 08:32:17 +0100, Harlequin [EMAIL PROTECTED] wrote: 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 $Query01 WHERE UserID='$_POST[TXT_UserID]' AND UserPassword='$_POST[TXT_UserPassword]'; $Result01 = mysql_query($Query01) or die(Error 01: . mysql_error()); [/CODE] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Query Returning Error
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 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
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.types.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
-
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PHP General Mailing List (http://www.php.net/)
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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:[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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RE: [PHP] Query Returning Error
To view the terms under which this email is distributed, please go to http://disclaimer.leedsmet.ac.uk/email.htm -Original Message- From: Chris Dowell [mailto:[EMAIL PROTECTED] Sent: 13 October 2004 11:57 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. Wrong -- the behaviour inside a quoted string is *more* correct, as it's a documented valid way to do it (whereas it's a documented way *not* to do it outside a string!). Cheers! Mike - Mike Ford, Electronic Information Services Adviser, Learning Support Services, Learning Information Services, JG125, James Graham Building, Leeds Metropolitan University, Headingley Campus, LEEDS, LS6 3QS, United Kingdom Email: [EMAIL PROTECTED] Tel: +44 113 283 2600 extn 4730 Fax: +44 113 283 3211 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
