Catcher.AndroidDemo.EasyRequestDemo.MainActivity.PostRequest C# (CSharp) Метод

        private async void PostRequest(object sender, EventArgs e)
        {
            string url = "http://192.168.1.102:8077/User/PostThing";
            IDictionary<string, string> routeParames = new Dictionary<string, string>();
            routeParames.Add("str", this.txtInput.Text);
            var result = await EasyWebRequest.SendPostRequestBasedOnHttpClient(url, routeParames);
            var data = (JsonObject)result;
            this.tv.Text = "hey," + data["Val"] + ",  i am from httpclient post";
        }