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REST - Hello World

Kashyap Hosdurga

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posted 8 years ago

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Hi,
I am learning REST and have written a simple hello world program.

package jerseyWebService;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;

// Plain old Java Object it does not extend as class or implements
// an interface

// The class registers its methods for the HTTP GET request using the @GET annotation.
// Using the @Produces annotation, it defines that it can deliver several MIME types,
// text, XML and HTML.

// The browser requests per default the HTML MIME type.

//Sets the path to base URL + /hello
@Path("/hello")
public class Hello {

// This method is called if TEXT_PLAIN is request
  @GET
  @Produces(MediaType.TEXT_PLAIN)
  public String sayPlainTextHello() {
    return "Hello Jersey";
  }

// This method is called if XML is request
  @GET
  @Produces(MediaType.TEXT_XML)
  public String sayXMLHello() {
    return "<?xml version=\"1.0\"?>" + "<hello> Hello Jersey" + "</hello>";
  }

// This method is called if HTML is request
  @GET
  @Produces(MediaType.TEXT_HTML)
  public String sayHtmlHello() {
    return "<html> " + "<title>" + "Hello Jersey" + "</title>"
        + "<body><h1>" + "Hello Jersey" + "</body></h1>" + "</html> ";
  }

}

My web.xml is

When I type
http://localhost:8080/ServletContainer1/test/hello

it says

type Status report

message /ServletContainer1/test/hello

description The requested resource is not available.

Can someone please help me to get this fixed?

Kashyap Hosdurga

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Posts: 98

posted 8 years ago

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Some more details

I am using Tomcat 8.0 and Eclipse Version: Neon.1a Release (4.6.1)

Tim Cooke

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posted 8 years ago

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Your URL mapping defines "/rest/*" but you are attempting to access "/test/hello".

Tim Driven Development | Test until the fear goes away

Kashyap Hosdurga

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posted 8 years ago

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Thanks for your response.

Even
http://localhost:8080/ServletContainer1/rest/hello

gives the same error message

Tim Moores

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Is the name of your web app really "ServletContainer1"? That doesn't sound like a name one would normally use.

If you renamed "com.vogella.jersey.first" to "test", then the URL would be http://localhost:8080/test/rest/hello

Kashyap Hosdurga

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posted 8 years ago

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No luck

, can you please guide me to some website that helps me with step by step instructions that get my first program working? I have gone thru a few websites (like vogella), but I am unable to find something that explains the link/connection between my servlet, XML file and the client/server (with tomcat/eclipse). Currently I am still reading "web services up and running" but wanted to make my hands dirty to understand the book better.

Gaston Greco

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posted 8 years ago

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I see you are using Jersey, you can explore the documentation on their website, there are many examples.
This page describes how to create a new REST API using a maven archetype, it should take no more than 5 minutes, you can then analyze the code while reading the docs: https://jersey.java.net/documentation/latest/getting-started.html

If later on you want to integrate Jersey with Spring and some data access, you can take a look at this boilerplate project https://github.com/gastongr/rest-boilerplate. You can get that running locally in 5 minutes too.

Also if you like video lessons, take a look at this great free course https://javabrains.io/courses/javaee_jaxrs

Consider Paul's rocket mass heater.