php - 如何在android中制作JSON响应的toast消息？

Question

• 点击删除按钮后，我想在 android 应用程序的 toast 中显示一条删除消息
• 任何人都可以告诉我下面的代码是正确的还是我必须在其中添加一些东西。

Android Java code

private void delete(){
    StringRequest stringRequest = new StringRequest(Request.Method.POST, "http://192.168.43.63/deletr.php",
            new Response.Listener<String>() {
                @Override
                public void onResponse(String response) {

                try {
                        JSONObject res = new JSONObject(response);
                    if (res.has("Deleted")) {
                        String aJsonString = res.getString("Deleted");
                        Toast.makeText(getActivity(), aJsonString, Toast.LENGTH_LONG).show();
                    }



                    } catch (JSONException e) {
                        e.printStackTrace();
                    }

                }
            },
            new Response.ErrorListener() {
                @Override
                public void onErrorResponse(VolleyError error) {
                    Toast.makeText(getActivity(), "Something went wrong",Toast.LENGTH_LONG).show();


                }
            }){
        @Override
        protected Map<String, String> getParams() throws AuthFailureError {
            Map<String, String> params = new HashMap<>();
            params.put("title",getCarEdit.getText().toString().trim());
            return params;
        }
    };
    RequestQueue requestQueue = Volley.newRequestQueue(getContext());
    requestQueue.add(stringRequest);

}

PHP code

 <?php
    include 'connection.php';

    global $connect;
    $id = $_POST["title"];


  $query = "DELETE FROM `products` WHERE title='$id'";

    $result = mysqli_query($connect, $query);

  if ($result) {

  echo json_encode("Deleted");
    }

    else{
       echo json_encode("Delete Failed");
    }

php response PHP response after asigned some manual values

Answer 1

PHP 文件中的更改:

$success = array( 
        "output"=>"your data Deleted"); 
$error = array( 
        "output"=>"your data not Deleted"); 

    // Use json_encode() function 
    if($result){
      $json = json_encode($success); 
    }else{
      $json = json_encode($error);
    }
    // Display the output 
    echo($json); 
    //
    // check hear your result if else code

    Output like : {"output":"your data Deleted"} 

在你的安卓代码中:

StringRequest request = new StringRequest(Request.Method.POST, URL, new Response.Listener<String>() {
      @Override
         public void onResponse(String response) {               
            Toast.makeText(MainActivity.this,response.toString(),Toast.LENGTH_LONG).show();
         }
      }, new Response.ErrorListener() {
         @Override
         public void onErrorResponse(VolleyError error) {
            Log.d("error",error.toString());
         }
      });

希望对您有所帮助...!