如何获取php中的json返回值? [英] how to grab the json returned value in php?

问题描述

我的Facebook应用程序中有一个脚本，该脚本将在下拉列表中为用户提供他的Facebook朋友列表.使用下面的代码，我可以抓住用户的朋友，但是我只获得用户ID.

I have a script in my facebook application which will provide the user with the list of his Facebook friends in a dropdown list. With the code below, I am able to grab the user's friends, but I am only getting the User ID.

$api_key = 'xxx';
$secret  = 'xxx';
include_once 'facebook.php';
$facebook = new Facebook($api_key, $secret);

$friends = $facebook->api_client->friends_get();
foreach ($friends as $friend)
{ 
  echo $friend;// returns only the friend's ID and not name but i want to show the name
  $url="https://graph.facebook.com/".$friend."/";
  $res=file_get_contents($url);
}

使用此链接，我想获取用户名并将其显示在下拉列表中.我该如何抓取它并仅回显$url中的名称?

With this link I want to grab the user's name and display it on a dropdown list. How can I grab it and echo only the name from $url?

{
   "id": "4",
   "name": "Mark Zuckerberg",
   "first_name": "Mark",
   "last_name": "Zuckerberg",
   "link": "http://www.facebook.com/zuck",
   "gender": "male",
   "locale": "en_US"
}

推荐答案

从PHP手册:

$var = json_decode ( $result );

并回显您想要的内容，例如:echo $var ['link'];

and echo what you want, like: echo $var ['link'];

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