How to get some values from a JSON string in C#?
A .NET 6 version using System.Text.Json
using System;
public class Program
{
public static void Main()
{
var jsonString = @"{ ""id"" : 123 }";
//parse it
var yourObject = System.Text.Json.JsonDocument.Parse(jsonString);
//retrieve the value
var id= yourObject.RootElement
.GetProperty("id");
Console.WriteLine(id);
}
}
To retrieve nested properties, you can chain the GetProperty
calls. As a more advanced example:
//access first element of array "persons" get nested property "age"
var age = yourObject.rootElement.GetProperty("persons")[0]
.GetProperty("age");
my string
var obj = {"Status":0,"Data":{"guid":"","invitationGuid":"","entityGuid":"387E22AD69-4910-430C-AC16-8044EE4A6B24443545DD"},"Extension":null}
Following code to get guid:
var userObj = JObject.Parse(obj);
var userGuid = Convert.ToString(userObj["Data"]["guid"]);
Your strings are JSON formatted, so you will need to parse it into a object. For that you can use JSON.NET.
Here is an example on how to parse a JSON string into a dynamic object:
string source = "{\r\n \"id\": \"100000280905615\", \r\n \"name\": \"Jerard Jones\", \r\n \"first_name\": \"Jerard\", \r\n \"last_name\": \"Jones\", \r\n \"link\": \"https://www.facebook.com/Jerard.Jones\", \r\n \"username\": \"Jerard.Jones\", \r\n \"gender\": \"female\", \r\n \"locale\": \"en_US\"\r\n}";
dynamic data = JObject.Parse(source);
Console.WriteLine(data.id);
Console.WriteLine(data.first_name);
Console.WriteLine(data.last_name);
Console.WriteLine(data.gender);
Console.WriteLine(data.locale);
Happy coding!
Following code is working for me.
Usings:
using System.IO;
using System.Net;
using Newtonsoft.Json.Linq;
Code:
using (HttpWebResponse response = (HttpWebResponse)request.GetResponse())
{
using (Stream responseStream = response.GetResponseStream())
{
using (StreamReader responseReader = new StreamReader(responseStream))
{
string json = responseReader.ReadToEnd();
string data = JObject.Parse(json)["id"].ToString();
}
}
}
//json = {"kind": "ALL", "id": "1221455", "longUrl": "NewURL"}