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PHP variable-infused link not writing to a variable?

开发者 https://www.devze.com 2023-04-11 20:16 出处:网络
http://www.reecemcmillin.com/albums/ <?php $uncut = file_get_contents(\'http://www.google.com/#sclient=psy-ab&hl=en&safe=active&source=hp&q=\' . $_POST[\'band\'] . \'+\' . $_POST[\

http://www.reecemcmillin.com/albums/

<?php
$uncut = file_get_contents('http://www.google.com/#sclient=psy-ab&hl=en&safe=active&source=hp&q=' . $_POST['band'] . '+' . $_POST['album'] . '+zip+inurl:mediafire');
$strip1 = strstr($uncut, 'www.mediafire.com/?');
$link = 开发者_JS百科substr($strip1, 0, 30);
echo $link;
?>

It doesn't seem to be writing the website content to $uncut. Can somebody help me figure out what's wrong? Thanks.<3


Clients are not supposed to send URI-fragments (the portion of the URI following #) to servers when they retrieve a document. PHP is probably sending a request for the google homepage, effectively: file_get_contents('http://www.google.com/');. If you echo $uncut, that's probably what you'll see you're getting back.

Try a querystring-based URI instead.

<?php
$uncut = file_get_contents('http://www.google.com/search?sclient=psy-ab&hl=en&safe=active&source=hp&q=' . urlencode($_POST['band']) . '+' . urlencode($_POST['album']) . '+zip+inurl:mediafire');
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