C# Post XML file using RestSharp

Question

I have written a method to post messages to an uri.

public string RestClientPost(string uri, string message = null)
    {
        var client = new RestClient(uri);
        var request = new RestRequest(Method.POST);
        request.AddHeader("Accept", "text/xml");
        if (!string.IsNullOrEmpty(message))
            request.AddParameter(message, ParameterType.RequestBody);

        var result = "";
        var response = client.Execute(request);

        if (response.StatusCode == HttpStatusCode.OK)
        {
            result = response.Content;
            Console.WriteLine(result);
        }
        else
        {
            result = response.StatusCode.ToString();
        }


        return result;
    }

and below code is used above method to post.

public void test123()
    {
        string uri = "myuri"; //private uri, cannot expose.
        var file= System.IO.File.ReadAllText(Path.Combine(Settings.EnvValPath, "RestClientXML", "test.XML"));
        var content = new RestClientServices().RestClientPost(uri, file);

    }

however, it returns "Unsupported Media type".

my test.XML's content is

<customer> 
    <customerName>test</customerName > 
    <customerStatus>OK</customerStatus > 
</customer>

And using Advanced Rest Client Plugin for Google Chrome, I'm able to post it and return with string that I wanted. Is there something wrong?? I set "content-type" to "text/xml" in Advanced Rest Client.

The return message is id of the customer. e.g: 2132

I change my original plan from RestSharp to Rest Client C# which I can get what I want now. If you guys need the answers, just drop me a message. — king jia
– king jia, Commented Jul 17, 2013 at 7:34

saber tabatabaee yazdi · Accepted Answer · 2019-01-15 14:08:11Z

1

im using postman,

if you can call any xml web services with this tools , then you can click on code and select restsharp and copy paste it to your code

answered Jan 15, 2019 at 14:08

saber tabatabaee yazdi

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Comments

bigge · Accepted Answer · 2013-07-17 05:16:35Z

0

This happened because the header "Accept" is to specify a type of return object. In this case a value of a variable content, not the type of content to send. Specify a type of content to send with: "Content-Type: application/xml".

If a return type of POST request is a media file, you can use 'image/png' or 'image/jpeg'. You can use multiple accept header values like: "application/xml, application/xhtml+xml, and image/png". For example, you can use Fiddler to debug HTTP(s) traffic - it's a good tool for web developers.

edited Jul 17, 2013 at 5:16

bigge

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answered Jul 17, 2013 at 4:01

Jose Rodriguez

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2 Comments

king jia Over a year ago

The return value is a string. E.g: 2312. Can you show an example? I don't really understand your answer.

Darius Over a year ago

What @Joseph wanted to say is that 'Accept' header indicates the type of response that you expect (see Accept Header. So in this specific case you don't need it You should specify request format: request.RequestFormat = DataFormat.Xml

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