How to call a Restful webservice in JAVA

Posted on 2013-11-21
Last Modified: 2013-12-11
 How do you call a RestFul WebService from a Java Application.  I need to call the URL and Post some data to it and retrive the response.

I was told that the service is written in JAVA and uses the Spring framework

Thanks for your help
Question by:SiemensSEN
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 3
  • 3
  • 2
LVL 28

Accepted Solution

dpearson earned 300 total points
ID: 39666703
Most people use a library like httpclient to handle this sort of web request.

There's details here on how to do a POST here:


(BTW it doesn't matter what the service itself is written in - to you it's just a URL that responds with some data).
LVL 36

Assisted Solution

mccarl earned 200 total points
ID: 39667759
If the webservice is fairly simple (in terms of the data that you send and receive, and perhaps the number of different types of calls/methods that you need to invoke) than you can probably use Doug's suggestions (or even just plain old Java HttpUrlConnection functionality) quite easily.

However, if the use of the webservice is more complex, you might want to investigate using a higher level library to help out. These libraries can handle automatic conversion of JSON/XML data that might be sent/received over the webservice, error handling, etc. There are many options out there but 2 that I have experience in and think are relatively simple to work with are:

JAX-RS on Java 6+ (Jersey Implementation) -
Spring -

Note I agree with Doug, that it doesn't matter what the webservice is written in. I mention Spring above because I think it is a good API to use for this sort of thing, NOT because the webservice uses it.

I know the above is a little bit brief, but if you decide that using a library such as these is the way to go, let us know and I can give you more detailed information.

Author Comment

ID: 39667961
Thank you .

I really will like to use spring. This RestAPI is simple now  but it will get more complicated in future releases.

I will read the links you sent me. However, if it would be very helpful  if you can provide any additional information. As always, I will Google to get more info

Thank you
Get 15 Days FREE Full-Featured Trial

Benefit from a mission critical IT monitoring with Monitis Premium or get it FREE for your entry level monitoring needs.
-Over 200,000 users
-More than 300,000 websites monitored
-Used in 197 countries
-Recommended by 98% of users


Author Comment

ID: 39668635
As an example.. I need to post the information below to the this rest URI

        "message" => "[Wed Jun 26 22:13:22 2013] [client] PHP Fatal   ...."
        ,"@timestamp" => "2013-06-27T02:13:22.000Z"
        ,"m_level"        => "Info"
        ,"m_cat"           => "THIRD_PARTY"
        ,"m_cor"           => "None"
        ,"m_tenant"      => "xxxxx"
        ,"m_timezone" => "EDT"
        ,"m_tzoffset"    => "-04:00"
        ,"m_host"         =>""
        ,"m_path"         => "/var/log/messages"
        ,"m_type"         => "linux-syslog"

I already constructed  the json object.
LVL 36

Assisted Solution

mccarl earned 200 total points
ID: 39668746
I will have some time later (it is the start of the weekend here and I don't get a lot of spare time over weekends to post on EE) to provide a more complete sample using the example you have given above, but to get you started it would be as simple as defining a bean in Spring for a RestTemplate object, and setting the list of messageConverters to use MappingJacksonHttpMessageConverter. You would probably also create a simple Java bean class to hold the fields in your message that you are sending, lets call it MessageDetail. Then you inject that restTemplate into your code and then the REST API call becomes as simple as...
MessageDetail message = new MessageDetail("[Wed Jun 26 22:13:22 2013] [error]...", "2013-06...", "Info", "THIRD_PARTY", ...);    // Note you don't even have to create the JSON yourself
URI location = restTemplate.postForLocation("", message);

Open in new window

Note that exactly which RestTemplate method you use depends on exactly how your REST service is implemented. The above assumes the REST standard where to add something you use the POST http method and the service returns a URI that describes where the newly added resource lives.

Author Comment

ID: 39670654
Thank you very much for the help. I will have a look at the RestTemplate and try to understand it.

However, I have to get a demo working this week. So, I will try use the HttpClient for now and later convert to REST.  Here is  my code snippet .. am I using the library correctly .  I will pass the JSON string and the URL to the class

public class Alarms 
     String url;
     public Alarms(String url)
          this.url = url;
     public void callRESTAPI(String jsonstring) throws IOException
         CloseableHttpResponse response=null;
         CloseableHttpClient httpclient = HttpClients.createDefault();
         HttpClient c = new HttpClient();
         HttpPost p = new HttpPost(this.url);
         p.setEntity(new StringEntity(jsonstring,ContentType.create("application/json")));
         response = httpclient.execute(p);		

Open in new window

LVL 28

Expert Comment

ID: 39670864
I haven't tried running your code - but it looks pretty good.

Fortunately httpclient is so widely used there's lots of examples on the web.  If you run into trouble a quick google search will show you lots of examples.

E.g. There's a whole tutorial here:

and this part has some more POST examples:

LVL 36

Expert Comment

ID: 39671316
And you can probably delete line 12 from the above code snippet. It's not doing anything useful.

Featured Post

Instantly Create Instructional Tutorials

Contextual Guidance at the moment of need helps your employees adopt to new software or processes instantly. Boost knowledge retention and employee engagement step-by-step with one easy solution.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Java had always been an easily readable and understandable language.  Some relatively recent changes in the language seem to be changing this pretty fast, and anyone that had not seen any Java code for the last 5 years will possibly have issues unde…
Java Flight Recorder and Java Mission Control together create a complete tool chain to continuously collect low level and detailed runtime information enabling after-the-fact incident analysis. Java Flight Recorder is a profiling and event collectio…
Viewers will learn one way to get user input in Java. Introduce the Scanner object: Declare the variable that stores the user input: An example prompting the user for input: Methods you need to invoke in order to properly get  user input:
Viewers will learn about if statements in Java and their use The if statement: The condition required to create an if statement: Variations of if statements: An example using if statements:

627 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question