N
N
Nikolai Gerasimov2016-02-06 23:26:11
PostgreSQL
Nikolai Gerasimov, 2016-02-06 23:26:11

Where to deploy a Java web application?

Wrote a small server in Java. I would like to try to make requests to it from the phone (android). So, are there hostings where you can deploy your server for FREE? The server itself is quite small, one war and a postgresql database. Run the server locally on Tomcat. It would be interesting to set up and run all this on the Internet.

Answer the question

In order to leave comments, you need to log in

9 answer(s)
E
Evgeny Maltsev, 2016-02-07
@hellpirat

https://www.heroku.com/

T
Timur, 2016-02-07
@timych

If the task is simple to test, then run it on a personal computer. In the configuration of the android application, specify your external ip and port. Only open it for external requests.

T
Timur Sergeevich, 2016-02-07
@MyAlesya

Get some cheap vps. Install java tomcat and deploy. If desired, fasten the domain (optional)

A
Anton Dzodzikov, 2016-02-07
@DzodzikovAK

Deploy this via Vagrant and connect to it as a remote machine. You will get an imitation of the fact that you are working with a project on a remote hosting.
By the way, Vagrant has Vagrant Share, which allows you to provide access to the container running on your computer from the outside (directly through the browser).

D
DKay, 2016-02-10
@DKay

Try a 14 day trial with Jelastic. Tomcat + PostgreSQL is deployed in a few minutes. Plus a few more minutes to deploy the application. Get a domain name too. The list of partners can be found here .

V
volyihin, 2016-03-10
@volyihin

Opebshift, Digitalocean. But it would be better/easier to emulate on your home computer.

R
Rad1calDreamer, 2014-10-02
@ArtMavir

Firstly, it is still correct to do this when forming a page in php.
Well, secondly, if you really need to do it with a script, then try like this

$(document).ready(function() {
      var link = window.location.pathname;
      $('.nav li.active').removeClass('active');
      $('.nav li a').each(function() {
          if ($(this).attr('href' == link) {
              $(this).parent('li').addClass('active');
            }
          });
      });
  });

V
Vitaly Kirenkov, 2014-10-01
@DeLaVega

Have you ever thought that it would be more correct to do this in php?

A
Azim Kurt, 2014-10-01
@Symphony

This is done in the backend

Didn't find what you were looking for?

Ask your question

Ask a Question

731 491 924 answers to any question