How to call the json key name (sample code) _ PHP Tutorial

The call method (sample code) for the json key name ). When the key name is a number or an abnormal variable character (if there is a space), you must use obj [xx] to obtain the value. Copy the code as follows :? Php declares json data $ arrayarray (res must use obj [xx] to obtain the value when the key name is a number or an abnormal variable character (such as a space.

The code is as follows:

// Declare json data
$ Array = array ('result' => array ("90" => "90 queue", "status" => "successful "));
$ Json = json_encode ($ array );

$ Array1 = array ("90" => "90 queue", "status" => "successful ");
$ Json1 = json_encode ($ array1 );
$ Phpjson = json_decode ($ json1, true); // The second parameter is true, which indicates converting json data into arrays.
// When the json key name is a number, you can only process $ phpjson ['90'] in array mode;
?>