php - 当php中的数组为空时，json_encode函数不返回大括号{}

Question

我有这个代码

$status = array(
                "message"=>"error",
                "club_id"=>$_club_id,
                "status"=>"1",
                "membership_info"=>array(),
                );

echo json_encode($status);

这个函数返回json:
{"message":"error","club_id":275,"status":"1","membership_info":[]}

但我需要这样的 json:

{"message":"error","club_id":275,"status":"1","membership_info":{}}

Answer 1

使用 json_encode 的 JSON_FORCE_OBJECT 选项:

json_encode($status, JSON_FORCE_OBJECT);

Documentation

JSON_FORCE_OBJECT (integer) Outputs an object rather than an array when a non-associative array is used. Especially useful when the recipient of the output is expecting an object and the array is empty. Available since PHP 5.3.0.

或者，如果你想在你的对象中保留你的“其他”数组，不要使用前面的答案，只需使用这个:

$status = array(
                "message"=>"error",
                "club_id"=>$_club_id,
                "status"=>"1",
                "membership_info"=> new stdClass()
                );