Twitter API 本地趋势解析 - stdClass 错误

发布于 2024-10-07 13:40:29 字数 1073 浏览 3 评论 0原文

我正在尝试解析某个位置（本例中为亚特兰大）的最新 Twitter 趋势，

这是我的代码：

<html>
  <head></head>
  <body>
    <?php
     $init = 'http://api.twitter.com/1/trends/2357024.json?count=1&callback=?&exclude=hashtags';
     $ch = curl_init();
     curl_setopt($ch, CURLOPT_URL,$init);
     curl_setopt($ch, CURLOPT_RETURNTRANSFER,1);
     $result = curl_exec($ch);
     curl_close($ch);
     $obj = json_decode($result);
     foreach ($obj[0]->trends as $trend) {
     echo "<li class=\"atlanta\">".$trend->name."</li>";
}?>
  </body>
</html>

所需的结果将是

Whatever The Trend Is

< /code>

它大约 30% 的时间有效 - 但其他 70% 我收到此错误：

Fatal error: Cannot use object of type stdClass as array in...

经过一番谷歌搜索后，似乎 obj 必须是数组...我找到的唯一答案是将 $obj 行更改为 true ，如下所示：

$obj = json_decode($result, true);

但是，这只会给我这个错误：

警告：为 foreach() 提供的参数无效...

有谁知道如何将我的代码更改为“数组”，以便它能够 100% 地工作？

原文

I'm trying to parse the latest Twitter trend from a location (Atlanta in this case)

Here's my code:

<html>
  <head></head>
  <body>
    <?php
     $init = 'http://api.twitter.com/1/trends/2357024.json?count=1&callback=?&exclude=hashtags';
     $ch = curl_init();
     curl_setopt($ch, CURLOPT_URL,$init);
     curl_setopt($ch, CURLOPT_RETURNTRANSFER,1);
     $result = curl_exec($ch);
     curl_close($ch);
     $obj = json_decode($result);
     foreach ($obj[0]->trends as $trend) {
     echo "<li class=\"atlanta\">".$trend->name."</li>";
}?>
  </body>
</html>

The desired result would be <li class="atlanta">Whatever The Trend Is</li>

It works about 30% of the time - but the other 70% I get this error:

Fatal error: Cannot use object of type stdClass as array in...

After some googling it seems the obj must be array... The only answer I've found is to change the $obj line to a true like this:

$obj = json_decode($result, true);

However, that simply gives me this error:

Warning: Invalid argument supplied for foreach() in...

Does anyone know how to change my code into an 'array' so that it will work 100% of the time?

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夏了南城 2024-10-14 13:40:29

@肯：试试

$obj = json_decode($result, true);
if (is_array($obj)) { 
    foreach ($obj[0]->trends as $trend) {
        echo "<li class=\"atlanta\">" . $trend->name . "</li>";
    }
} else {
   // failure case here
}

@Ken: Try

$obj = json_decode($result, true);
if (is_array($obj)) { 
    foreach ($obj[0]->trends as $trend) {
        echo "<li class=\"atlanta\">" . $trend->name . "</li>";
    }
} else {
   // failure case here
}

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