PHP中的$ _POST没有使用Volley库JsonObjectRequest（方法POST）填充

Question

Let's start with the code I am using, I have tried every single possible different way to make "params". I have used it as a HashMap, in Json format and also as a string. I have also tried to @Override the getParams() method by creating a hashmap and returning it. Nothing as worked.

Here is my function that calls the JsonObjectRequest.

 private void sendJsonObjReq() {

    showProgressDialog();
    Map<String, String> para = new HashMap<>();

    para.put("Condicao", "dea");
    para.put("Field", "1550");
    JSONObject jason = new JSONObject(para);

    String params = jason.toString();
    textView.setText(params);

        JsonObjectRequest jsonReq = new JsonObjectRequest(JsonRequest.Method.POST,
                url_cond1, params,
                new Response.Listener<JSONObject>() {

                    @Override
                    public void onResponse(JSONObject response) {
                        Log.d(AppController.TAG, response.toString());
                        textView.setText(response.toString());

                    /*try {
                        int u = response.getInt("sucess");
                        if(u == 1){

                            JSONArray json = response.optJSONArray("Type");
                            if(json != null) {
                                MakeListHashMap(json);
                            }else{
                                textView.setText("Wrong Parameters");
                            }
                        }else{textView.setText("success is 0");}
                    }catch(JSONException e){
                        Log.e("Error:", e.getMessage());
                        textView.setText("nothing here");
                    }*/
                        hideProgressDialog();
                    }
                }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {
                VolleyLog.d(AppController.TAG, "Error:" + error.getMessage());
                showProgressDialog();
            }
        });
        //Add to request queue
    AppController.getInstance().addToRequestQueue(jsonReq, tag_json_obj);

}

The url and everything else is fine, i have checked but i just cannot understand why neither the GET or POST method work, since i have tried both but I did not have any success with them. My php code consists of this:

<?php
    $response = array();
//$oi = json_decode($_POST);
//$response["Condicao"] = $_GET["Condicao"];
//$response["Field"] = $_GET["Field"];
    $response["Condicao"] = $_POST["Condicao"];
    $response["Field"] = $_POST["Field"];

    $response['sucess'] = 1;
    $response['other'] = "test";

    echo json_encode($response);
?>

I have tried decoding it and not decoding it, I really am at a loss for what to do. I think it might be a problem with the server but using the App "Postman" I can send GET and POST and the response is a Json array like i want.

If i send key1= Condicao => name=dea; key2= Field => name=1550;

I get the result

{
     "Condicao": "dea",
     "Field": "1550",
     "sucess": 1,
     "other": "test"
}

edit: The solution is:

<?php
    $_POST = json_decode(file_get_contents('php://input'), true);
    $response = array();

    $response["Condicao"] = $_POST["Condicao"];
    $response["Field"] = $_POST["Field"];

    $response['sucess'] = 1;
    $response['other'] = "test";

    echo json_encode($response);
?>

Answer 1

1.) Pass your json object instead of string

JsonObjectRequest jsonReq = new JsonObjectRequest(JsonRequest.Method.POST,
                url_cond1, jason ,
               //          ^^^^
                new Response.Listener<JSONObject>() {

2.) Receive it in your php as

<?php
    $response   = array();
    // receive your json object
    $jasonarray = json_decode(file_get_contents('php://input'),true);
    $response["Condicao"] = $jasonarray["Condicao"];
    $response["Field"]    = $jasonarray["Field"];
    $response['sucess']   = 1;
    $response['other']    = "test";

    echo json_encode($response);
?>