CODEWITHSUNDEEP
php
Sammy Miri
Sammy Miri

Reputation: 123

How can I remove invalid querystring using php header location

I have this invalid link hard coded in software which I cannot modify.

http://www.16start.com/results.php?cof=GALT:#FFFFFF;GL:1;DIV:#FFFFFF;FORID:1&q=search

I would like to use php header location to redirect it to a valid URL which does not contain the querystring. I'd like to pass just the parameter q=.

I've tried 

$q = $_GET['q'];
header ("Location: http://www.newURL.com/results.php?" . $q . "");

But it's just passing the invalid querystring to the new location in addition to modifying it in a strange way

This is the destination location I get, which is also invalid

http://www.newURL.com/results.php?#FFFFFF;GL:1;DIV:#FFFFFF;FORID:1&q=search

Upvotes: 4

Views: 760

Answers (2)

huysentruitw
huysentruitw

Reputation: 28141

That's because # is seen as the start of a fragment identifier and confuses the parser.

You can take the easy-way as Stretch suggested but you should be aware that q is the last query parameter in your URL. Therefore, it might be better to fix the URL and extract the query parameters in a safer way:

<?php
$url = "http://www.16start.com/results.php?cof=GALT:#FFFFFF;GL:1;DIV:#FFFFFF;FORID:1&q=search";

// Replace # with its HTML entity:
$url = str_replace('#', "%23", $url);

// Extract the query part from the URL
$query = parse_url($url, PHP_URL_QUERY);

// From here on you could prepend the new url
$newUrl = "http://www.newURL.com/results.php?" . $query;
var_dump($newUrl);

// Or you can even go further and convert the query part into an array
parse_str($query, $params);
var_dump($params);
?>

Output

string 'http://www.newURL.com/results.php?cof=GALT:%23FFFFFF;GL:1;DIV:%23FFFFFF;FORID:1&q=search' (length=88)

array
  'cof' => string 'GALT:#FFFFFF;GL:1;DIV:#FFFFFF;FORID:1' (length=37)
  'q' => string 'search' (length=6)

Update

After your comments, it seems that the URL is not available as a string in your script and you want to get it from the browser.

The bad news is that PHP will not receive the fragment part (everything after the #), because it is not sent to the server. You can verify this if you check the network tab in the Development tools of your browser F12.

In this case, you'll have to host a page at http://www.16start.com/results.php that contains some client-side JavaScript for parsing the fragment and redirecting the user.

Upvotes: 1

Just Lucky Really
Just Lucky Really

Reputation: 1401

one way could be to use strstr() to get everything after (and including q=) in the string.

So:

$q=strstr($_GET['q'],'q=');

Give that a whirl

Upvotes: 0

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