Php 在这个场景中读取JSON？

我试图解析来自不同来源的rss提要，其中一个来源是：

http://feeds.feedburner.com/DiscoveryNews-Top-Stories

但是这个源代码给了我一些奇怪的json数据，比如：

"item": [
    {
     "title": [
      "Snazzy Science Photos of the Week (August 10-16)",
      {
       "type": "html",
       "content": "Snazzy Science Photos of the Week (August 10-16)"
      }
     ],
     "description": [
      "Glowing rabbits, treasure-hunting badgers and a case of mistaken UFO identity help round out this week&#039;s photos.<img src=\"http://feeds.feedburner.com/~r/DiscoveryNews-Top-Stories/~4/S6Urfvdw2DQ\" height=\"1\" width=\"1\"/>",
      {
       "type": "html",
       "content": "Glowing rabbits, treasure-hunting badgers and a case of mistaken UFO identity help round out this week&#039;s photos."
      }
     ],

<?php
$url = 'http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20feed%20where%20url=%22http://feeds.feedburner.com/DiscoveryNews-Top-Stories%22&format=json&diagnostics=true&callback=cbfunc';
$data = substr(file_get_contents($url), 7, -2);
$json = json_decode($data);
foreach ($json->query->results->item as $item)
{
    echo "Title: ", $item->title[0], "\nDescription: ", $item->description[0], "\n";
    echo "==================================================\n\n";
}

但是当我试图解析Discovery新闻提要时，上面的代码失败了。请帮助

[编辑]：

---

我使用YQL从源代码获取等效的JSON。这是它将元素打包为数组的

"title": [
          "No Battery Required for This Wireless Device",
          {
              "type": "html",
              "content": "No Battery Required for This Wireless Device"
          }
         ],

您可以这样读取第一个元素：

"item": [
    {
     "title": [
      "Snazzy Science Photos of the Week (August 10-16)",
      {
       "type": "html",
       "content": "Snazzy Science Photos of the Week (August 10-16)"
      }
     ],
     "description": [
      "Glowing rabbits, treasure-hunting badgers and a case of mistaken UFO identity help round out this week&#039;s photos.<img src=\"http://feeds.feedburner.com/~r/DiscoveryNews-Top-Stories/~4/S6Urfvdw2DQ\" height=\"1\" width=\"1\"/>",
      {
       "type": "html",
       "content": "Glowing rabbits, treasure-hunting badgers and a case of mistaken UFO identity help round out this week&#039;s photos."
      }
     ],

<?php
$url = 'http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20feed%20where%20url=%22http://feeds.feedburner.com/DiscoveryNews-Top-Stories%22&format=json&diagnostics=true&callback=cbfunc';
$data = substr(file_get_contents($url), 7, -2);
$json = json_decode($data);
foreach ($json->query->results->item as $item)
{
    echo "Title: ", $item->title[0], "\nDescription: ", $item->description[0], "\n";
    echo "==================================================\n\n";
}

---

我没有看到任何JSON，只有XML可用。@RajatSaxena你能展示（JSFIDLE？）你是如何使用YQL/JSON提要的吗？我只是从PHP解析它，我没有使用javascript。另外，我只粘贴了我在问题中使用的PHP代码。谢谢，这有助于解决问题lot@RajatSaxena哈哈：P把我的坏消息贴出来成了一种习惯。顺便问一下，您不想直接读取XML有什么原因吗？不，我似乎喜欢JSON。@RajatSaxena如果您感兴趣的话，我已经在底部发布了一个简单的示例，说明如何使用SimpleXML读取它；）您知道如何在php中确定字符串是文本还是html吗？