A simple question

zingbats

I have a script that logs you in. But, I want it to redirect you to the page you logged in at.

in the header.location bit of the code, what code could I use that names the page that sent me to login.php (pref. full URL)?

so.. you login at test1.php then the code in login.php redirects you back to test1.php

Cheers Alex

cgraz

header("Location: test1.php");

Cgraz

zingbats

I know how to do that.. I mean...

say you went to login.php via test1.php, when you login, it would redirect you to test1.php, but, if you logged in on test2.php, it would redirect you to test2.php etc

I need to make a $tag that is what the refer page was.

so.. it would be

header("Location: $refer");

regards;
Alex.

mystrymaster

I imagine there is a simpler way to do it but you could have the redirect to login work like

then have the login script grab $_GET['where'];

zingbats

yeah, but that STILL doesn't redirect them from where they came from.

m_tt

header("Location: " . $_SERVER['PHP_SELF']);

zingbats

That actually doesn't work.. firstly because it should be:

header("Location: " . $_SERVER['PHP_SELF'] . " ");

And secondly it doesn't work.

I tried

<?
$serverthing = "$_SERVER['PHP_SELF']";
?>

Then echoed $serverthing and it gives an error.

zingbats

also,

<?
$serverthing = $_SERVER['PHP_SELF'];
?>

Gives the name of the page that you are actually ON, not the redirect

stolzyboy

have you tried HTTP_REFERER

zingbats

no, please explain

stolzyboy

$_SERVER['HTTP_REFERER'] // will give you the referring page, so if you grab this variable at login.php, you should get say page2.php, if you went there and it sent you to login.php, you may want to echo it out to see what it holds while you are testing

sidenote, you wouldn't want to use this as a security option for say only letting certain people in from certain urls, it is spoofable, but who cares on a simple redirect if it is spoofable

zingbats

<?
$refer = $_SERVER['HTTP_REFERER'];
?>
<? echo ("$refer");?>

prints nothing

stolzyboy

you put it on the login.php page and tried to go to some other page that redirected you to it?

cuz it should work, i have used it this same situation

m_tt

Originally posted by zingbats
That actually doesn't work.. firstly because it should be:
header("Location: " . $_SERVER['PHP_SELF'] . " ");
And secondly it doesn't work.

I tried

<?
$serverthing = "$_SERVER['PHP_SELF']";
?>

Then echoed $serverthing and it gives an error. [/B]

Firstly, you're wrong. Secondly, you're calling a variable as a string "".

If you want to go back to the previous page it is possible to use HTTP_REFERER depending on how your form/redirect works after login. Should be something like

header("Location: " . $_SERVER['HTTP_REFERER']);
header("Location: " . getenv('HTTP_REFERER'));

If you redirect from a POST then the headers will be empty and the HTTP_REFERER value will be blank.

feldon23

Why have a separate login page anyway?

stolzyboy

he has a seperate login, cuz he has checks on certain pages that you need to be logged in at, and instead having login code on every page, he checks for a session, if there isn't one, he redirects them to the login page

pskZero7

Hello zingbats, try this, I think it will work fine :

header("location:".$_SERVER[HTTP_REFERER]);

NOTA : $SERVER the new name of $HTTP_SERVER_VARS. It's been introduced in PHP 4.1.0 so if you're using an older version you should probably use $HTTP_SERVER_VARS instead of $SERVER

zingbats

Nope. Still can't get it to work

I have tried:

<?
$referthingy = getenv('HTTP_REFERER');
?>

<? echo ("$referthingy");?>

------------------
and

<?
$referthingy = $_SERVER[HTTP_REFERER];
?>

<? echo ("$referthingy");?>

Also, if I do the header thingy, it just redirects me to the homepage (http://www.tycooneden.com/)

stolzyboy

what version of php are you using??

this is confusing why it doesn't work

benson

Hello,
I had the same problem a while ago.
What i did was with a mail function.
I had a form and a hidden variable.
The hidden variable was $ref

in the php post file.
I say
Header("location: $ref ")

I dunno if this help you but it works