RestSharp deserialize XML node inside a parent node

Question

I'm calling an API service that wraps all responses surrounded by a Response element:

<?xml version="1.0" encoding="UTF-8"?>
<Response xmlns="http://www.example.com/stuff">
  <status upstairs="Normal" downstairs="Normal"/>
</Response>

My status class is:

class status {
    public string upstairs { get; set; }
    public string downstairs { get; set; }
}

I'm calling RestSharp to get the status (and prior to V106 this worked):

var client = new RestClient("http://www.example.com");
var request = new RestRequest("/api?get=status", Method.Get);
var response = client.Execute<status>(request);
if (response.Status.Equals(OK))
{
    var status = response.Data;
}

With the current version of RestSharp I see in response.Content that the data is there, however the deserializer failed with error

"<Response xmlns="http://www.example.com/stuff"> was not expected."

How do I specify that there is a parent element, but to ignore it?

erict · Accepted Answer · 2024-05-08 21:10:19Z

0

Got it to work by changing my model to include the outer element, like so:

[XmlRoot("Response", Namespace="http://www.example.com/stuff")]
public class ResponseStatus {
    [XmlElement]
    public status status { get; set; }
}

public class status {
    [XmlAttribute]
    public string upstairs { get; set; }
    [XmlAttribute]
    public string downstairs { get; set; }
}

The code now looks like this:

var client = new RestClient("http://www.example.com");
var request = new RestRequest("/api?get=status", Method.Get);
var response = client.Execute<ResponseStatus>(request);
if (response.Status.Equals(OK)) {
    var status = response.Data.status;
}

answered May 8, 2024 at 21:10

erict

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