Question

i have a RESTful WCF service and one of its methods use an Object as parameter

[WebInvoke(UriTemplate = "save", Method = "POST", RequestFormat = WebMessageFormat.Xml, ResponseFormat= WebMessageFormat.Xml), OperationContract]
        public SampleItem Create(SampleItem instance)
        {
            return new SampleItem() { Id = 1, StringValue = "saved" };
            // TODO: Add the new instance of SampleItem to the collection
            //throw new NotImplementedException();
        }

I am trying to call this method from my eclipse android project. i am using these lines of codes

DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost post=new HttpPost("http://10.0.2.2:2768/Service1.svc/save");
ArrayList<NameValuePair> nvp= new ArrayList<NameValuePair>();

nvp.add(new BasicNameValuePair("Id", "1"));
nvp.add(new BasicNameValuePair("StringValue", "yolo"));

post.setEntity(new UrlEncodedFormEntity(nvp));
HttpResponse httpResponse = httpClient.execute(post);
HttpEntity httpEntity = httpResponse.getEntity();
String xml = EntityUtils.toString(httpEntity);

Every time i get this error Method not allowed. in the XML that is returned by the service method.

i have tried invoking it from the browser, but encountered the same error there.

please tell me what i am doing wrong and what i can do instead.

thanks in advance to anyone who can help.

note: other methods which do not use object as parameter are working fine.

EDIT: tried Fiddler2 with success. but stalled again.

i have tried invoking the method SampleItem Create(SampleItem instance) with the url http://localhost:2768/Service1.svc/save and it works. the method returns the object in XML format.

in fiddler i added the request body as
<SampleItem xmlns="http://schemas.datacontract.org/2004/07/WcfRestService1" xmlns:i="http://www.w3.org/2001/XMLSchema-instance"><Id>1</Id><StringValue>saved</StringValue></SampleItem>

but the problem is that i can not find any way to add this xml string to the HttpPost or HttpRequest as the requestbody eclipse android project.

note: passing the xml string as Header or UrlEncodedFormEntity did not work.

Answer 1

First, you should get the Web Service method working from the browser – I recommend using Fiddler2 – its easier to construct the request body with your object and also to set the request headers when doing a post. It will show you the response so should help with debugging.
As for your code, I’m doing a POST to a WCF service and instead of doing

post.setEntity(new UrlEncodedFormEntity(nvp));

I’m simply doing:

HttpPost request = new HttpPost(url);

// Add headers.
for(NameValuePair h : headers)
{
     request.addHeader(h.getName(), h.getValue());
}

(I am using JSONObjects and I have RequestFormat = WebMessageFormat.Json in my WebInvoke parameters.

Also, check your using the correct UriTemplate name in your url as they are case sensitive.