View source с PHP

Автор: talismana

<head><meta http-equiv="Content-Type" content="text/html; charset=windows-1251" /></head>
<center>
<form action="?" method="GET">

<input type="text" name="url">
<br>
<input type="submit" value="Show source!">
</form>
</center>
<HR>
<?php
error_reporting(0);
show_source($_GET['url']);
?>