_*I am behind proxy and i tried to decode *JSON* encoded script behind

Question

I am behind proxy  and  i tried to  decode JSON* encoded script behind proxy, Using curl i read the url and then feed that to JSON_DECODE but this is giving*
invalid argument or error at JSON_DECODE can u expalin any solution?
<?php
session_start();
require_once('JSON.phps');
$url = 'http://ajax.googleapis.com/ajax/services/search/web?rsz=large&v=1.0&q=' . urlencode('site:' . $_SERVER['localhost'] . ' ' . $_POST['searchquery']);
$ch = curl_init();
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 0);
curl_setopt($ch, CURLOPT_PROXY, "172.31.1.1");
curl_setopt($ch, CURLOPT_PROXYPORT, 8080);
curl_setopt($ch, CURLOPT_PROXYUSERPWD, "myid:myPASS");
curl_setopt($ch, CURLOPT_URL, $url);
$body = curl_exec($ch) or die(curl_error($ch));
curl_close($ch);
$json = new Services_JSON();
$json = $json->decode($body);   //Giving error at this point as INVALID ARGUMENT
foreach($json->responseData->results as $searchresult)
if($searchresult->GsearchResultClass == 'GwebSearch')
$formattedresults .= '
<div class="searchresult">
<h3><a href="' . $searchresult->unescapedUrl . '">' . $searchresult->titleNoFormatting . '</a></h3>
<p class="resultdesc">' . $searchresult->content . '</p>
<p class="resulturl">' . $searchresult->visibleUrl . '</p>
</div>';
$_SESSION['googleresults'] = $formattedresults;
header('Location: ' . $_SERVER['HTTP_REFERER']);
exit;
?>

Answer 1

This is a Java forum. Your script appears to be written in PHP.
Please note that Javascript and Java are unrelated.