Question

任何人都可以帮我解决这个错误吗？一直在寻找它，但一直没能让它工作。我刚才写了这篇文章，当我上次离开它的时候感觉它很有用，但是当我再次访问它时我无法运行它。感谢您的帮助。

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：第46行的/Applications/MAMP/htdocs/search/views/layouts/search.php中为foreach（）提供的参数无效

第46行是最后一行代码的第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')
{

Answer 1

可能它返回了一个空的回复。在使用foreach之前，使用is_object($json->responseData)测试变量很有用。我还会检查属性'结果'是否与property_exists($json, 'results')

一起存在

注意：尝试在第46行的... / search.php中获取非对象的属性

1 个答案: