How to trim $_SERVER['HTTP_REFERER']

Question

After from processing i am sending the user on the previous page using:

header('Location: ' . $_SERVER['HTTP_REFERER'] . '?Add=Successful');

Now it sends me to my add.php:

http://localhost/add.php?Add=Successful

Again when i add one more data the header location passes the following:

http://localhost/add.php?Add=Successful?Add=Successful

What i want is to trim the header location till question mark:

Lets say something like trimming the $_SERVER['HTTP_REFERER'] till ? and saving it into a variable so that if keyword ? exists it should trim it again to http://localhost/add.php and then pass that variable into header location, so that it can become something like this:

header('Location: ' . $trimmedHeader . '?Add=Successful');

Naki · Accepted Answer · 2014-01-03T11:25:21.580

9

You can also use PHP parse_url() function.

$url = parse_url($_SERVER['HTTP_REFERER']);
$trimmedHeader = $url['scheme'] . '://' . $url['host'] . $url['path'];
header('Location: ' . $trimmedHeader . '?Add=Successful');

edited Jan 03 '14 at 11:25

answered Mar 29 '12 at 12:27

Naki

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All of you have given the answer and its working, but the problem is, its not passing the variable back along with header. Its passing this:- `http://localhost/add.php` whereas i want to pass this:- `http://localhost/add.php?Add=Successful` again. – Django Anonymous Mar 29 '12 at 12:43
But this what is in my script pass this variable again: '?Add=Successful'. What more do You need? – Naki Mar 29 '12 at 12:55

kappa · Answer 2 · 2012-08-05T10:07:09.797

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preg_replace('/(.*)\?/',$_SERVER['HTTP_REFERER'],'\1');

edited Aug 05 '12 at 10:07

answered Mar 29 '12 at 12:17

kappa

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score 0 · Answer 3 · answered Mar 29 '12 at 12:25

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This will return you everything before the first question mark in the string.

$trimmedheader = array_shift(explode("?", $_SERVER['HTTP_REFERER']));

answered Mar 29 '12 at 12:25

Pierre-Olivier

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All of you have given the answer and its working, but the problem is, its not passing the variable back along with header. Its passing this:- `http://localhost/add.php` whereas i want to pass this:- `http://localhost/add.php?Add=Successful` again. – Django Anonymous Mar 29 '12 at 12:43
You said: `What i want is to trim the header location till question mark` That is what my script is doing. If you want to append "?Add=Successful" to your string, just add it at the end: `array_shift(explode("?", $_SERVER['HTTP_REFERER'])) . "?Add=Successful"` – Pierre-Olivier Mar 29 '12 at 12:49

Ingmar Boddington · Answer 4 · 2012-03-29T13:43:14.483

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$urlArray = parse_url($_SERVER['HTTP_REFERER']);
$newUrl = $urlArray['scheme'].'://'.$urlArray['host'].$urlArray['path'].'?Add=Successful';
header("Location: $newUrl");

This has been tested and works fine....

edited Mar 29 '12 at 13:43

answered Mar 29 '12 at 12:29

Ingmar Boddington

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All of you have given the answer and its working, but the problem is, its not passing the variable back along with header. Its passing this:- `http://localhost/add.php` whereas i want to pass this:- `http://localhost/add.php?Add=Successful` again. – Django Anonymous Mar 29 '12 at 12:42
Try concatenating outside the header line maybe (I know function returns in header strings foofs things up) – Ingmar Boddington Mar 29 '12 at 13:41

How to trim $_SERVER['HTTP_REFERER']

4 Answers4