谁能帮我解决这个错误？一直在看它，但一直无法让它工作。前一阵子写了这个，上次离开时感觉它可以工作，但是当我重新访问它时，我一直无法运行它。感谢所有的帮助。
注意：尝试在第 46 行的 /Applications/MAMP/htdocs/search/views/layouts/search.php 中获取非对象的属性
警告：在第 46 行的 /Applications/MAMP/htdocs/search/views/layouts/search.php 中为 foreach() 提供的参数无效
第 46 行是 foreach 代码最后一行的第 4 行。
<?php
include($_SERVER['DOCUMENT_ROOT'].'/'.'search/scripts/JSON.php');
include($_SERVER['DOCUMENT_ROOT'].'/'.'search/scripts/search_fns.php');
$searchquery = urlencode(isset($_GET['search']) ? $_GET['search'] : "news");
// Google Search API
$url = "http://ajax.googleapis.com/ajax/services/search/web?v=1.0&"
. "q="
. $searchquery
."&key=ABQIAAAAYIeqEnf9yNjBzcHJK7yDdhSklBzi76D_F0lniPI7JR27aK7eCBSU-xpNs1axVS45y_PX_7_ibsScUA&userip=USERS-IP-ADDRESS&rsz=filtered_cse";
// sendRequest
// note how referer is set manually
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_REFERER, "http://localhost");
$body = curl_exec($ch);
curl_close($ch);
// process the JSON string
$json = new Services_JSON;
$json = $json->decode($body);
$formattedresults = "";
$rating = "";
$search = $searchquery;
?>
<style type="text/css">
td img {display: block;}
</style>
<div id="main">
<?php
foreach($json->responseData->results as $searchresult)
{
if($searchresult->GsearchResultClass == 'GwebSearch')
{