Question 247260 in Json Question posted in
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finding specific Json field value without model class

I am looking a way to get the specific value from complex json without creating the Model class. I am able to read via Newtonsoft but with model and since I have many different json model,is it possible to get the value of content.result.id directly ?
Is ther any way to get it as my json keeps on changing:
Json:

{
    "id": "123",
    "Type": "Test",
    "content": {
        "id": "abcd",
        "testResult": "Failed",
        "field1": {
            "id": "T1",
            "name": "T1 name"
        },
        "filed2": {
            "id": "T2",
            "name": "T2 name "
        },
        "result": {
            "id": "1111",
            "name": "T3 name"
        }
    },
    "createdAt": "2018-10-11T13:18:54.0555456+01:00",
    "updatedAt": "2018-10-11T13:18:54.0555518+01:00",
    "deleted": false,
    "_rid": "AAAA===",
    "_self": "Ab++/",
    "_etag": ""1qw"",
    "_attachments": "attachments/",
    "_ts": 12345
}

Expected outcome: 1111 (from within result)

Tags:

4

Answers


  1. 0

    With Newtonsoft.Json, read the JSON as JObject and get the value with JSON Path.

    using Newtonsoft.Json.Linq;

JObject jObj = JObject.Parse(json);
string id = jObj.SelectToken("$.content.result.id")
    .ToString();

    Reference: Querying JSON with complex JSON Path

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  2. 0

    if you are using System.Text.Json you can use JsonNode as Follows :

    using System.Text.Json.Nodes;

var jsonObject = JsonNode.Parse(json)!.AsObject();

var id = jsonObject["content"]?["result"]?["id"];

    or if you are using Newtonsoft you can use JObject as Follows:

    using Newtonsoft.Json.Linq;
    
var jsonObject = JObject.Parse(json);

var id = jsonObject["content"]?["result"]?["id"]?.ToString();
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  3. 0 
    var parsed = JObject.Parse(json);
var resultId = parsed["content"]["result"]["id"];

    You need to add null checking, but that will work.

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  4. 0

    You can fetch the specific JSON field value without defining a model class by using Newtonsoft’s JObject class

    …

    using Newtonsoft.Json.Linq;

string json = @"{
// JSON here
}";

JObject jsonObj = JObject.Parse(json);
string idValue = (string)jsonObj["content"]["result"]["id"];
Console.WriteLine("The ID value is: " + idValue);

    …

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