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/"
   }
}

PHP from error below:

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

Error:

PHP Notice: Undefined property: stdClass::$comments in /mysite/public_html/wp-content/themes/theme/functions.php on line 797

My partial fix:

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

Answer 1

You have to wrap the $url in curly braces:

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