Question

I am successfully created restful web service and deploy it in Apache Tomcat 7.0. After successful deployment I start my server. By using the below command i invoke the web service.

WebResource resource = client.resource("http://localhost:8080/rest/samp/create"); 

My web method is

@POST
@Path("/create")
@Produces(MediaType.TEXT_XML)
@Consumes(MediaType.TEXT_XML)
public final String sample(final String xmlMessage) {

    return "<xml version=1.0><welcome>"+xmlmessage+"</welcome>";    
}

Here I am passing XML content as argument and get the XML content as response.

Now what I need is how to pass the XML content to the web method.

Answer 1

I’m guessing that the library in use here is Jersey.

You have to use a builder to set the appropriate HTTP headers, method and entity body.

WebResource resource = client.resource("http://localhost:8080/rest/samp/create");
String request = "<your_xml>...</your_xml>";
String response = resource.accept(
     MediaType.TEXT_XML).
     header("X-FOO", "BAR"). //this line is not necessary, just an example
     type(MediaType.TEXT_XML).
     post(String.class, request);

But I recommend using JAXB instead. Creating XML as plain strings is just crude and unnecessarily annoying. It doesn’t show in such a simple example (grabbing a whole XML and wrapping it with another tag) but it will soon enough.

This tutorial should get you started.