在PHP中解析Facebook JSON时遇到问题

我正试图从Facebook页面上调出帖子，但我遇到了一些问题。我在解析Twitter提要的JSON时没有问题，所以我看不出我遇到了什么问题

这是我的代码，还有：

<?php
    $url = "http://www.facebook.com/feeds/page.php?format=json&id=96551536516";             

    function disguise_curl($url) { 
        $curl = curl_init(); 
        $header[0] = "Accept: text/xml,application/xml,application/xhtml+xml,"; 
        $header[0] .= "text/html;q=0.9,text/plain;q=0.8,image/png,*/*;q=0.5"; 
        $header[] = "Cache-Control: max-age=0"; 
        $header[] = "Connection: keep-alive"; 
        $header[] = "Keep-Alive: 300"; 
        $header[] = "Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.7"; 
        $header[] = "Accept-Language: en-us,en;q=0.5"; 
        $header[] = "Pragma: ";

        curl_setopt($curl, CURLOPT_URL, $url); 
        curl_setopt($curl, CURLOPT_USERAGENT, 'Mozilla'); 
        curl_setopt($curl, CURLOPT_HTTPHEADER, $header); 
        curl_setopt($curl, CURLOPT_REFERER, ''); 
        curl_setopt($curl, CURLOPT_ENCODING, 'gzip,deflate'); 
        curl_setopt($curl, CURLOPT_AUTOREFERER, true); 
        curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1); 
        curl_setopt($curl, CURLOPT_TIMEOUT, 10); 

        $html = curl_exec($curl); 
        curl_close($curl);

        return $html;
    }

    $response = json_decode(disguise_curl($url));

    foreach($response->entries as $block){
        echo
            "<li class='clearfix'>
                <div class='streamPosterName'>{$block->author}</div>
                <div class='postContent'>{$block->title}</div>
            </li>";
    }
?>

条目作为$block）{
回声
“

{$block->author}
{$block->title}
”；
}
?>

在页面的其他部分解析JSON时，我使用以下方法引用JSON对象：

<?php
    $url = "http://search.twitter.com/search.json?q=Apple&rpp=50";             
    $ch = curl_init(); 
    curl_setopt($ch, CURLOPT_URL, $url);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
    $curlout = curl_exec($ch);
    curl_close($ch);
    $response = json_decode($curlout, true);

    foreach($response["results"] as $block){
        echo
            "<li class='clearfix'>
                <img src='".$block["profile_image_url"]."' />
                <div class='streamPosterName'>".$block["from_user_name"]."</div>
                <div class='streamPosterUsername'>@".$block["from_user"]."</div>
                <div class='postContent'>".$block["text"]."</div>
            </li>";
    }
?>

{$block->author}

默认情况下，json_decode（）将json对象表示为PHP stdClass对象。该错误意味着
$block->author
是一个对象，但您使用它的方式就像使用字符串一样

也许可以试试
$block->author->name
？
如果您定义了“不工作”，那么它会很有帮助。这只是意味着$block->author是Object类型而不是String，因此不是Echoableah dang，愚蠢的错误。没有完全读取JSON对象。谢谢
<div class='streamPosterName'>{$block->author}</div>

<div class='streamPosterName'>{$block->author}</div>