Swagger PHP：如何声明属性以使用模式定义？

Question

My App's response looks like this:

{
  "status": "success",
  "data": {
      "status": "ready"
   },
  "request_id": "string"
}

I tried to define response in Swagger

 *           @SWG\Response (
 *              response=200,
 *              description="Success response",
 *              @SWG\Schema (
 *                  @SWG\Property(
 *                      property="status",
 *                      type="string",
 *                      default="success"
 *                  ),
 *                  @SWG\Property(
 *                      property="data",
 *                      @SWG\Schema(
 *                          ref="#/definitions/Service/models/Status"
 *                      )
 *                  ),
 *                  @SWG\Property(
 *                      property="request_id",
 *                      type="string"
 *                  ),
 *              )
 *          ),

But it does not use Schema definition for Status, so my response actually looks like:

{
  "status": "success",
  "data": {},
  "request_id": "string"
}

How can I define data property to use schema definition? Or can it can be done in a different way?

Answer 1

How funny can be the fact, that people sometimes find the answer just after posting a question.

Answer is:

*           @SWG\Response (
 *              response=200,
 *              description="Success response",
 *              @SWG\Schema (
 *                  @SWG\Property(
 *                      property="status",
 *                      type="string",
 *                      default="success"
 *                  ),
 *                  @SWG\Property(
 *                      property="data",
 *                      ref="#/definitions/Service/models/Status"
 *                  ),
 *                  @SWG\Property(
 *                      property="request_id",
 *                      type="string"
 *                  ),
 *              )
 *          ),