How can I parse_url in PHP when there is a URL in a string variable?
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by Eric O
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Published on 2010-04-23T19:14:13Z
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2010/04/23
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I am admittedly a PHP newbie, so I need some help.
I am creating a self-designed affiliate program for my site and have the option for an affiliate to add a SubID to their link for tracking. Without having control over what is entered, I have been testing different scenarios and found a bug when a full URL is entered (i.e. "http://example.com").
In my PHP I can grab the variable from the string no problem. My problem comes from when I get the referring URL and parse it (since I need to parse the referring URL to get the host mane for other uses). Code below:
$refURL = getenv("HTTP_REFERER");
$parseRefURL = parse_url($refURL);
WORKS when incoming link is (for example):
http://example.com/?ref=REFERRER'S-ID&sid=www.test.com
ERROR when incoming link is (notice the addition of "http://" after "sid="):
http://example.com/?ref=REFERRER'S-ID&sid=http://www.test.com
Here is the warning message:
Warning: parse_url(/?ref=REFERRER'S-ID&sid=http://www.test.com) [function.parse-url]: Unable to parse url in /home4/'directory'/public_html/hosterdoodle/header.php on line 28
Any ideas on how to keep the parse-url function from being thrown off when someone may decide to place a URL in a variable? (I actually tested this problem down to the point that it will throw the error with as little as ":/" in the variable)
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