Correct JSON structure using RestSharp

Question

What is the Correct format for sending JSON using RestSharp:

Example PUT JSON:
 {
 "properties": [
{
  "name": "description",
  "value": "A far better description than before"
}
]
}

In C# how to correctly send, I'm attempting with:

     request.AddJsonBody(new
        {
            properties = new[]
            {
                new{property="name",value="about_us"},
                new{property="value",value="My description"}
            }
        });

Below is the full code:

  private void UpdateCompanyProperty(string companyId)
    {

        var hapikey = "{YOUR_HAPI_KEY_HERE}";
        var client = new RestClient("https://api.hubapi.com/");
        var request = new RestRequest("companies/v2/companies/{companyId}", Method.PUT);
        request.AddUrlSegment("companyId", companyId);
        request.AddQueryParameter("hapikey", hapikey);
        request.RequestFormat = DataFormat.Json;

        request.AddJsonBody(new
        {
            properties = new[]
            {
                new{property="name",value="about_us"},
                new{property="value",value="My description"}
            }
        });

        IRestResponse response = client.Execute(request);

        JObject jObject = JObject.Parse(response.Content);
        JToken jvid = jObject["portalId"];

        Debug.WriteLine(jvid);

    }

No errors but not updating or returning values.

score 0 · Accepted Answer · answered Jul 30 '19 at 23:14

0

Try my answer here:

https://stackoverflow.com/a/57281157/5478655

request.RequestFormat = DataFormat.Json; // Important

var input = new Dictionary<string, object>();
// props could be an array or real objects too of course
var props = new[]
{
    new{property="name",value="about_us"},
    new{property="value",value="My description"}
};
input.Add("properties", props);

request.AddBody(input);

answered Jul 30 '19 at 23:14

LucasMetal

1,323
10
16

Thank you, I also added a response if anyone else is looking. – Zealous Sep 26 '19 at 23:24

score 0 · Answer 2 · answered Feb 24 '21 at 12:01

Create a class and give it any name

class MyClass
{
   public string property {get;set;}
   private string value {get;set;}
}

Define your class as an object

List<MyClass> list = new List<MyClass>
{
   new MyClass() { property = "name", value = "about_us"},
   new MyClass() { property = "value", value = "My Description"},
};

Now using Newtonsoft.Json serialize your object

string result = JsonConvert.SerializeObject(list);

Now add it to an array

 var resArray = new object[] { result };

Find your modified code below

class MyClass
    {
       public string property {get;set;}
       private string value {get;set;}
    }

using Newtonsoft.Json;
using RestSharp;

private void UpdateCompanyProperty(string companyId)
{


List<MyClass> list = new List<MyClass>
    {
       new MyClass() { property = "name", value = "about_us"},
       new MyClass() { property = "value", value = "My Description"},
    };

string result = JsonConvert.SerializeObject(list);

    var hapikey = "{YOUR_HAPI_KEY_HERE}";
    var client = new RestClient("https://api.hubapi.com/");
    var request = new RestRequest("companies/v2/companies/{companyId}", Method.PUT);
    request.AddUrlSegment("companyId", companyId);
    request.AddQueryParameter("hapikey", hapikey);
    request.RequestFormat = DataFormat.Json;

    request.AddJsonBody(new
    {
        properties =result
    });

    IRestResponse response = client.Execute(request);

    JObject jObject = JObject.Parse(response.Content);
    JToken jvid = jObject["portalId"];

    Debug.WriteLine(jvid);

}

Correct JSON structure using RestSharp

2 Answers2