Deserializing XML Response using RestSharp

Question

I have read through the various questions already posed here regarding this subject and i'm still no closer to solving my problem.

I am trying to Deserialize this xml response:

 <?xml version="1.0" encoding="ISO-8859-1" ?>
   <SubmissionResult>
     <Result>ACCEPTED</Result>
       <SubmissionID>
         <RefNo>77587-1425386500</RefNo>
         <Submitted>9</Submitted>
         <ValidNo>7</ValidNo>
         <InvalidNo>2</InvalidNo>
         <InvalidNo_1>08452782055</InvalidNo_1>
         <InvalidNo_2>01234567890</InvalidNo_1>
         <TextvID>77587-1425386500</TextvID>
       </SubmissionID>
     <Credits>999999</Credits>
   </SubmissionResult>

Using these classes:

[XmlRoot ("SubmissionResult")]
public class SubmissionResult
{
    [XmlElement ("Result")]
    public string Result { get; set; }
    public SubmissionID SubmissionID { get; set; }
    [XmlElement("Credits")]
    public int Credits { get; set; }
}

public class SubmissionID
{
    [XmlElement("RefNo")]    
    public int RefNo { get; set; }
    [XmlElement("Submitted")]    
    public int Submitted { get; set; }
    [XmlElement("ValidNo")]    
    public int ValidNo { get; set; }
    [XmlElement("TextVID")]    
    public string TextVID { get; set; }
}

However I am only ever returning null and 0 values, which I assume are the default.

Here is the code to get the results and hopefully deserialize:

try
{
    var request = new RestRequest();
    request.RequestFormat = DataFormat.Xml;
    request.Resource = APIURL;
    request.RootElement = "SubmissionResult";

    SubmissionResult r = Execute<SubmissionResult>(request);
}

public static T Execute<T>(RestRequest request) where T : new()
{
     var client = new RestClient();
         client.BaseUrl = new  Uri("https://www.textvertising.co.uk/_admin/api", UriKind.Absolute);

     var response = client.Execute<T>(request);

     return response.Data;
}

Many thanks in advance for any help you can provide.

CS

TextvID has incompatible casing in the XML and code attribute. Minor and probably not relevant, but you should still fix that. — Marcel, Aug 03 '17 at 09:04

score 5 · Accepted Answer · answered Feb 24 '16 at 10:49

I made a few changes and could make it work:

Probably a typo in the question but something I had to change is that posted XML is invalid:

<InvalidNo_2>01234567890</InvalidNo_1>

I understand you want use .NET XMLSerializer as you used XmlRoot and XmlElement annotations so you have to override the default one that comes with RestSharp like so:

request.XmlSerializer = new RestSharp.Serializers.DotNetXmlSerializer();

Also I had to delete the RootElement setting which didn't play nicely with .NET serializer so deleted the following line:

request.RootElement = "SubmissionResult";

So my version became:

var request = new RestRequest();
request.XmlSerializer = new RestSharp.Serializers.DotNetXmlSerializer();
request.RequestFormat = DataFormat.Xml;
// request.Resource = APIURL;

SubmissionResult r = Execute<SubmissionResult>(request);

Finally I noticed RefNo is meant to be integer in the class but the returned value is not (77587-1425386500) so I made it a string:

[XmlElement("RefNo")]
public string RefNo { get; set; }

I created a mock response at mocky.io at tested with that and it seems to be working fine:

public static T Execute<T>(RestRequest request) where T : new()
{
    var client = new RestClient();
    client.BaseUrl = new Uri("http://www.mocky.io/v2/56cd88660f00009800d61ff8", UriKind.Absolute);

    var response = client.Execute<T>(request);

    return response.Data;
}

score 4 · Answer 2 · answered Jun 15 '21 at 11:03

Simple code to De-serialize xml response

//Create object of RestClient
IRestClient restClient = new RestClient();

//Create request with return type of payload as xml and method as GET
IRestRequest restRequest = new RestRequest(Url);
restRequest.AddHeader("Accept", "application/xml");
restRequest.Method = Method.GET;

//Execute the request to fetch the response
IRestResponse restResponse = restClient.Execute(restRequest);

//This is wrapper for System.Xml.Serialization.XmlSerializer 
var dotNetXmlDeserializer = new RestSharp.Deserializers.DotNetXmlDeserializer();
ModelClassName modelClassObject =
 dotNetXmlDeserializer.Deserialize<ModelClassName>(restResponse);

 Console.WriteLine(ModelClassName.PropertyName);

Where ModelClassName is the POCO class we create to store the response in de-serialized form.

You can paste the copied payload in JSON/XML format into POCO class by using Visual Studios in built feature. Edit -> Paste Special -> Paste JSON/XML as classes

"Edit -> Paste Special -> Paste JSON/XML as classes" Wow, I wished I knew about this a decade ago! Excellent tip :) — Radderz, Aug 17 '22 at 20:24

score -3 · Answer 3 · answered Feb 24 '16 at 10:19

Using Restsharp i normally do a different approach but i am mainly working with JSON. Is this using MVC web api?

I would create a Client being of Type IRestClient where i would set the API url and some headers. I would then use,

var response = client.Post<T>(DataFormat.Json, "admin/api", content); 
if(response.Data != null)
//populate T object with data

If you could please provide some more information on the web service method you are calling.. it would help thanks :) good luck with this..

Have a look at this http://restsharp.org/ To be honest i have never used the RestClient.Execute<> myself but this link should hopefully help

Deserializing XML Response using RestSharp

3 Answers3

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