RESTful Java client with java.net.URL
Posted on , Last modified : August 29, 2012
By mkyong
In this tutorial, we show you how to create a RESTful Java client with Java build-in HTTP client library. It’s simple to use and good enough to perform basic operations for REST service.
The RESTful services from last “Jackson + JAX-RS” article will be reused, and we will use “java.net.URL” and “java.net.HttpURLConnection” to create a simple Java client to send “GET” and “POST” request.
1. GET Request
Review last REST service, return “json” data back to client.
@Path("/json/product") public class JSONService { @GET @Path("/get") @Produces("application/json") public Product getProductInJSON() { Product product = new Product(); product.setName("iPad 3"); product.setQty(999); return product; } //...
Java client to send a “GET” request.
import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.net.HttpURLConnection; import java.net.MalformedURLException; import java.net.URL; public class NetClientGet { // http://localhost:8080/RESTfulExample/json/product/get public static void main(String[] args) { try { URL url = new URL("http://localhost:8080/RESTfulExample/json/product/get"); HttpURLConnection conn = (HttpURLConnection) url.openConnection(); conn.setRequestMethod("GET"); conn.setRequestProperty("Accept", "application/json"); if (conn.getResponseCode() != 200) { throw new RuntimeException("Failed : HTTP error code : " + conn.getResponseCode()); } BufferedReader br = new BufferedReader(new InputStreamReader( (conn.getInputStream()))); String output; System.out.println("Output from Server .... \n"); while ((output = br.readLine()) != null) { System.out.println(output); } conn.disconnect(); } catch (MalformedURLException e) { e.printStackTrace(); } catch (IOException e) { e.printStackTrace(); } } }
Output…
Output from Server .... {"qty":999,"name":"iPad 3"}
2. POST Request
Review last REST service, accept “json” data and convert it into Product object, via Jackson provider automatically.
@Path("/json/product") public class JSONService { @POST @Path("/post") @Consumes("application/json") public Response createProductInJSON(Product product) { String result = "Product created : " + product; return Response.status(201).entity(result).build(); } //...
Java client to send a “POST” request, with json string.
import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.OutputStream; import java.net.HttpURLConnection; import java.net.MalformedURLException; import java.net.URL; public class NetClientPost { // http://localhost:8080/RESTfulExample/json/product/post public static void main(String[] args) { try { URL url = new URL("http://localhost:8080/RESTfulExample/json/product/post"); HttpURLConnection conn = (HttpURLConnection) url.openConnection(); conn.setDoOutput(true); conn.setRequestMethod("POST"); conn.setRequestProperty("Content-Type", "application/json"); String input = "{\"qty\":100,\"name\":\"iPad 4\"}"; OutputStream os = conn.getOutputStream(); os.write(input.getBytes()); os.flush(); if (conn.getResponseCode() != HttpURLConnection.HTTP_CREATED) { throw new RuntimeException("Failed : HTTP error code : " + conn.getResponseCode()); } BufferedReader br = new BufferedReader(new InputStreamReader( (conn.getInputStream()))); String output; System.out.println("Output from Server .... \n"); while ((output = br.readLine()) != null) { System.out.println(output); } conn.disconnect(); } catch (MalformedURLException e) { e.printStackTrace(); } catch (IOException e) { e.printStackTrace(); } } }
Output…
Output from Server .... Product created : Product [name=iPad 4, qty=100]
Download Source Code
Download it – JAX-RS-Client-JavaURL-Example.zip (8 KB)
References
Here are some of my recommended Books
Thanks mkyong.
As a Java programmer completely new to json, this article provided all the information I needed to understand how to request data from a json API.
Since the service I’m connecting to is https, the only major changes I needed to make were:
substitute
import javax.net.ssl.HttpsURLConnection;
for
import java.net.HttpURLConnection;
and, in the code:
HttpsURLConnection conn = (HttpsURLConnection) url.openConnection();
instead of
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
I also had to change:
to
otherwise the exception was thrown even when the connection had been correctly established.
to
instead of
Great article.
I am unable to form the request parameter for consuming REST web service GET and POST, could any one please guide based on below scenario.
If I have REST webservice as below which is expecting two string parameters
General Web URL to consume webservice :
http://localhost:8080/myWs/sayHello?name=Peter&msg=Hai
//How to pass the arguments for getting GET and POST result.
org.springframework.web.client.RestTemplate restTemplate = new RestTemplate();
String url = “http://localhost:8080/myWs/sayHello”;
Map vars = new HashMap();
vars.put(“name”, “peter”);
vars.put(“msg”, “Hai”);
String result = restTemplate.getForObject(url+”/{name}/{msg}”, String.class, vars);
String result1 = restTemplate.postForObject(url, vars,String.class);
System.out.println(“GET result : “+result + “\nPOST result1″+result1);
Great article. I used this to create my Java client to talk to a .NET WebAPI REST service that I created. Works like a charm.
Can you also add a PUT and DELETE snippet to make this post a complete tutorial?