Question

我有来自谷歌的这个脚本它工作得很好，代码能够从任何页面检索任何Facebook相册突然它停止了工作我在谷歌上做了一项研究我发现它需要在代码中实现访问令牌但我不能单独做任何帮助请问候

$fb_page_id = "448955931844556";

$json_link = "http://graph.facebook.com/{$fb_page_id}/albums?fields=id,name,description,link,cover_photo,count";
$json = file_get_contents($json_link);

$obj = json_decode($json, true, 512, JSON_BIGINT_AS_STRING);

$album_count = count($obj['data']);
for($x=0; $x<$album_count; $x++){

    $id = $obj['data'][$x]['id'];
    $name = $obj['data'][$x]['name'];
    $description = $obj['data'][$x]['description'];
    $link = $obj['data'][$x]['link'];
    $cover_photo = $obj['data'][$x]['cover_photo'];
    $count = $obj['data'][$x]['count'];

    // if you want to exclude an album, just add the name on the if statement
    if(
        $name!="Profile Pictures" &&
        $name!="Cover Photos" &&
        $name!="Timeline Photos"
    ){

        $show_pictures_link = "photos.php?album_id={$id}&album_name={$name}";

        echo "<div class='col-md-4'>";
            echo "<a href='{$show_pictures_link}'>";
                echo "<img class='img-responsive' style='height:190px;width:190px;' src='https://graph.facebook.com/{$cover_photo}/picture' alt=''>";
            echo "</a>";
            echo "<h3>";
                echo "<a href='{$show_pictures_link}'>{$name}</a>";
            echo "</h3>";

            $count_text="Photo";
            if($count>1){ $count_text="Photos"; }

            echo "<p>";
                echo "<div style='color:#888;'>{$count} {$count_text} / <a href='{$link}' target='_blank'>View on Facebook</a></div>";
                 $description;
            echo "</p>";
        echo "</div>";
    }
}

Answer 1

将access_token=<your access token>添加到$json_link

例如：

$fb_access_token = "PUT YOUR ACCESS TOKEN HERE" 
$json_link = "http://graph.facebook.com/{$fb_page_id}/albums?access_token={$fb_access_token}&fields=id,name,description,link,cover_photo,count";

将fb访问令牌插入此代码

1 个答案: