将JSON索引作为PHP变量访问

Question

$ JSON：

{
   "https://google.com/": {
      "share": {
         "comments": 10,
         "shares": 20
      },
      "id": "https://google.com/"
   }
}

以下错误的PHP：

$url = "https://google.com/";
... json is fetched here and set as $json
$count = $json->$url->comments; 

错误：

  

PHP注意：第797行/mysite/public_html/wp-content/themes/theme/functions.php中的未定义属性：stdClass :: $ comments

我的部分修复：

$count = $json->$url->share->comments;

Answer 1

你必须将$url包裹在花括号中：

$count = $json->{$url}->share->comments;

Answer 2

首先尝试解码json。

$json = json_decode($json);

$url = "https://google.com/";

$count = $json->$url->share->comments;