RESTful call in Java

JavaRest

Java Problem Overview


I am going to make a RESTful call in Java. However, I don't know how to make the call. Do I need to use the URLConnection or others?

Java Solutions


Solution 1 - Java

Update: It’s been almost 5 years since I wrote the answer below; today I have a different perspective.

99% of the time when people use the term REST, they really mean HTTP; they could care less about “resources”, “representations”, “state transfers”, “uniform interfaces”, “hypermedia”, or any other constraints or aspects of the REST architecture style identified by Fielding. The abstractions provided by various REST frameworks are therefore confusing and unhelpful.

So: you want to send HTTP requests using Java in 2015. You want an API that is clear, expressive, intuitive, idiomatic, simple. What to use? I no longer use Java, but for the past few years the Java HTTP client library that has seemed the most promising and interesting is OkHttp. Check it out.


You can definitely interact with RESTful web services by using URLConnection or HTTPClient to code HTTP requests.

However, it's generally more desirable to use a library or framework which provides a simpler and more semantic API specifically designed for this purpose. This makes the code easier to write, read, and debug, and reduces duplication of effort. These frameworks generally implement some great features which aren't necessarily present or easy to use in lower-level libraries, such as content negotiation, caching, and authentication.

Some of the most mature options are Jersey, RESTEasy, and Restlet.

I'm most familiar with Restlet, and Jersey, let's look at how we'd make a POST request with both APIs.

Jersey Example

Form form = new Form();
form.add("x", "foo");
form.add("y", "bar");

Client client = ClientBuilder.newClient();
    
WebTarget resource = client.target("http://localhost:8080/someresource");

Builder request = resource.request();
request.accept(MediaType.APPLICATION_JSON);

Response response = request.get();

if (response.getStatusInfo().getFamily() == Family.SUCCESSFUL) {
    System.out.println("Success! " + response.getStatus());
    System.out.println(response.getEntity());
} else {
    System.out.println("ERROR! " + response.getStatus());    
    System.out.println(response.getEntity());
}

Restlet Example

Form form = new Form();
form.add("x", "foo");
form.add("y", "bar");

ClientResource resource = new ClientResource("http://localhost:8080/someresource");

Response response = resource.post(form.getWebRepresentation());

if (response.getStatus().isSuccess()) {
    System.out.println("Success! " + response.getStatus());
    System.out.println(response.getEntity().getText());
} else {
    System.out.println("ERROR! " + response.getStatus());
    System.out.println(response.getEntity().getText());
}

Of course, GET requests are even simpler, and you can also specify things like entity tags and Accept headers, but hopefully these examples are usefully non-trivial but not too complex.

As you can see, Restlet and Jersey have similar client APIs. I believe they were developed around the same time, and therefore influenced each other.

I find the Restlet API to be a little more semantic, and therefore a little clearer, but YMMV.

As I said, I'm most familiar with Restlet, I've used it in many apps for years, and I'm very happy with it. It's a very mature, robust, simple, effective, active, and well-supported framework. I can't speak to Jersey or RESTEasy, but my impression is that they're both also solid choices.

Solution 2 - Java

If you are calling a RESTful service from a Service Provider (e.g Facebook, Twitter), you can do it with any flavour of your choice:

If you don't want to use external libraries, you can use java.net.HttpURLConnection or javax.net.ssl.HttpsURLConnection (for SSL), but that is call encapsulated in a Factory type pattern in java.net.URLConnection. To receive the result, you will have to connection.getInputStream() which returns you an InputStream. You will then have to convert your input stream to string and parse the string into it's representative object (e.g. XML, JSON, etc).

Alternatively, Apache HttpClient (version 4 is the latest). It's more stable and robust than java's default URLConnection and it supports most (if not all) HTTP protocol (as well as it can be set to Strict mode). Your response will still be in InputStream and you can use it as mentioned above.


Documentation on HttpClient: http://hc.apache.org/httpcomponents-client-ga/tutorial/html/index.html

Solution 3 - Java

This is very complicated in java, which is why I would suggest using Spring's RestTemplate abstraction:

String result = 
restTemplate.getForObject(
    "http://example.com/hotels/{hotel}/bookings/{booking}",
    String.class,"42", "21"
);

Reference:

Solution 4 - Java

If you just need to make a simple call to a REST service from java you use something along these line

/*
 * Stolen from http://xml.nig.ac.jp/tutorial/rest/index.html
 * and http://www.dr-chuck.com/csev-blog/2007/09/calling-rest-web-services-from-java/
*/
import java.io.*;
import java.net.*;

public class Rest {

    public static void main(String[] args) throws IOException {
        URL url = new URL(INSERT_HERE_YOUR_URL);
        String query = INSERT_HERE_YOUR_URL_PARAMETERS;

        //make connection
        URLConnection urlc = url.openConnection();

        //use post mode
        urlc.setDoOutput(true);
        urlc.setAllowUserInteraction(false);

        //send query
        PrintStream ps = new PrintStream(urlc.getOutputStream());
        ps.print(query);
        ps.close();

        //get result
        BufferedReader br = new BufferedReader(new InputStreamReader(urlc
            .getInputStream()));
        String l = null;
        while ((l=br.readLine())!=null) {
            System.out.println(l);
        }
        br.close();
	}
}

Solution 5 - Java

There are several RESTful APIs around. I would recommend Jersey;

https://jersey.java.net/

Client API documentation is here;

https://jersey.java.net/documentation/latest/index.html

Update
Location for the OAuth docs in the comment below is a dead link and has moved to https://jersey.java.net/nonav/documentation/latest/security.html#d0e12334

Solution 6 - Java

I want to share my personal experience, calling a REST WS with Post JSON call:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.net.URL;
import java.net.URLConnection;

public class HWS {

    public static void main(String[] args) throws IOException {
        URL url = new URL("INSERT YOUR SERVER REQUEST");
        //Insert your JSON query request
        String query = "{'PARAM1': 'VALUE','PARAM2': 'VALUE','PARAM3': 'VALUE','PARAM4': 'VALUE'}";
        //It change the apostrophe char to double quote char, to form a correct JSON string
        query=query.replace("'", "\"");

        try{
	        //make connection
	        URLConnection urlc = url.openConnection();
	        //It Content Type is so important to support JSON call
	        urlc.setRequestProperty("Content-Type", "application/xml");
	        Msj("Conectando: " + url.toString());
	        //use post mode
	        urlc.setDoOutput(true);
	        urlc.setAllowUserInteraction(false);
	
	        //send query
	        PrintStream ps = new PrintStream(urlc.getOutputStream());
	        ps.print(query);
	        Msj("Consulta: " + query);
	        ps.close();
	
	        //get result
	        BufferedReader br = new BufferedReader(new InputStreamReader(urlc.getInputStream()));
	        String l = null;
	        while ((l=br.readLine())!=null) {
	            Msj(l);
	        }
	        br.close();
        } catch (Exception e){
        	Msj("Error ocurrido");
        	Msj(e.toString());
        }
    }
	
	private static void Msj(String texto){
    	System.out.println(texto);
    }
}

Solution 7 - Java

You can check out the CXF. You can visit the JAX-RS Article here

Calling is as simple as this (quote):

BookStore store = JAXRSClientFactory.create("http://bookstore.com", BookStore.class);
// (1) remote GET call to http://bookstore.com/bookstore
Books books = store.getAllBooks();
// (2) no remote call
BookResource subresource = store.getBookSubresource(1);
// {3} remote GET call to http://bookstore.com/bookstore/1
Book b = subresource.getDescription();

Solution 8 - Java

Most Easy Solution will be using Apache http client library. refer following sample code.. this code uses basic security for authenticating.

Add following Dependency.

> > org.apache.httpcomponents > httpclient > 4.4 >

CredentialsProvider credentialsProvider = new BasicCredentialsProvider();
Credentials credentials = new UsernamePasswordCredentials("username", "password");
credentialsProvider.setCredentials(AuthScope.ANY, credentials);
HttpClient client = HttpClientBuilder.create().setDefaultCredentialsProvider(credentialsProvider).build();
HttpPost request = new HttpPost("https://api.plivo.com/v1/Account/MAYNJ3OT/Message/");HttpResponse response = client.execute(request);
    // Get the response
    BufferedReader rd = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
    String line = "";
    while ((line = rd.readLine()) != null) {    
    textView = textView + line;
    }
    System.out.println(textView);
        		

Solution 9 - Java

Indeed, this is "very complicated in Java":

From: https://jersey.java.net/documentation/latest/client.html

Client client = ClientBuilder.newClient();
WebTarget target = client.target("http://foo").path("bar");
Invocation.Builder invocationBuilder = target.request(MediaType.TEXT_PLAIN_TYPE);
Response response = invocationBuilder.get();

Solution 10 - Java

I see many answers, here's what we are using in 2020 WebClient, and BTW RestTemplate is going to get deprecated. (can check this) RestTemplate going to be deprecated

Solution 11 - Java

The following is an example of a GET request that prints the response body as a String with HttpRequest:

   HttpClient client = HttpClient.newHttpClient();
   HttpRequest request = HttpRequest.newBuilder()
         .uri(URI.create("http://example.org/"))
         .build();
   client.sendAsync(request, BodyHandlers.ofString())
         .thenApply(HttpResponse::body)
         .thenAccept(System.out::println)
         .join(); 

If you need to set the header you could use the following method inside the HttpRequest.Builder:

.setHeader("api-key", "value")

Solution 12 - Java

You can use Async Http Client (The library also supports the WebSocket Protocol) like that:

	String clientChannel = UriBuilder.fromPath("http://localhost:8080/api/{id}").build(id).toString();

	try (AsyncHttpClient asyncHttpClient = new AsyncHttpClient())
	{
		BoundRequestBuilder postRequest = asyncHttpClient.preparePost(clientChannel);
		postRequest.setHeader(HttpHeaders.CONTENT_TYPE, MediaType.APPLICATION_JSON);
		postRequest.setBody(message.toString()); // returns JSON
		postRequest.execute().get();
	}

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionQuestionsView Question on Stackoverflow
Solution 1 - JavaAvi FlaxView Answer on Stackoverflow
Solution 2 - JavaBuhake SindiView Answer on Stackoverflow
Solution 3 - JavaSean Patrick FloydView Answer on Stackoverflow
Solution 4 - JavajitterView Answer on Stackoverflow
Solution 5 - JavaQwerkyView Answer on Stackoverflow
Solution 6 - JavaLisandroView Answer on Stackoverflow
Solution 7 - JavaJoopiterView Answer on Stackoverflow
Solution 8 - JavaHeminView Answer on Stackoverflow
Solution 9 - JavaJRunView Answer on Stackoverflow
Solution 10 - JavaFrancis RajView Answer on Stackoverflow
Solution 11 - JavaG. CiardiniView Answer on Stackoverflow
Solution 12 - Javakayz1View Answer on Stackoverflow