CODEWITHSUNDEEP
php
Carl Johnson
Carl Johnson

Reputation: 9

PHP Harmful URL protection

I've made this script, but the 4th line isn't right and I have really no clue how to solve this. I really appriciate if someone helps me. This is my code:

<?php
$url = $_GET["url"];
$badsite = array("http://check.com", "http://hotmail.com");

if($url == $badsite) {
echo "This URL is harmful.";
} else {
echo "Not harmful";
header("Location: " . $_GET["url"]);
}
?>

So the thing which doesn't work is the following line

if($url == $badsite) {

How can I make it so it checks if the GET contains a $badsite?

Upvotes: 0

Views: 118

Answers (2)

Adrian Cid Almaguer
Adrian Cid Almaguer

Reputation: 7791

In this case you can use the function in_array:

http://php.net/manual/en/function.in-array.php

<?php
$url = $_GET["url"];

$badsite = array("http://check.com", "http://hotmail.com");

if(in_array($url, $basite)) {
echo "This URL is harmful.";
} else {
echo "Not harmful";
header("Location: " . $_GET["url"]);
}
?>

Upvotes: 0

David
David

Reputation: 218798

You don't want to check if the value equals the array, you want to check if it's in the array. Perhaps something like this:

if (in_array($url, $badsite)) {
  // ...
}

Side note, you don't need (or want, really) this echo statement:

echo "Not harmful";
header("Location: " . $_GET["url"]);

You might get an error by emitting output before sending a header. But even if you buffer output or in some other way suppress that error, there's no reason to emit output when returning a redirect response. The browser would display it for only an instant, if at all. A redirect by itself is a complete HTTP response, no output is required.

Upvotes: 1

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