REST Web Services 11 - Setting Up
Why Stack Overflow recommends it
Stack Overflow users link it as the tutorial they followed to create a Jersey Maven project, usually when their setup then fails with build errors, missing libraries or 404s.
Recommended by 1 people across 1 different months, 2015 to 2020.
What it is
A short Java Brains tutorial from the JAX-RS REST Web Services series that walks through creating a new Maven project in Eclipse using the Jersey quickstart webapp archetype. It sets up the environment for building a REST API on Tomcat. Presented by Koushik Kothagal of Java Brains.
What people said
Recommends benz · 2015-05 · Stack Overflow · on “Jersey: The request resource is not available”
I followed the guide of the guy in the tutorial which I want to work through [this video] I created it from eclipse maven but I will try to create once from the context menu. Thanks for your support.
Cites Rahil Kumar · 2016-05 · Stack Overflow · on “Setiing up jersy Restful maven Project in Eclipse”
Today I am started with Restful Practice , I got Stuck in Setting up Restful maven project . I am creating a Maven project,In Select Archetype window i am not getting any option related to jersey in filter text box during search of jersey API... Steps I Followed after watching Below Video Tutorial : [this video] Click on New option ---> other Search for MAven and click on Maven Project Click next---- again click on next In Filter text box I am searching for "Jersey", (I am not getting options related to jerseyglassfish, I am expecting one option on search name "com.glassfish.jersey.archetypes" containing two different 2 artifact ID is this need any plugin Installation or library attachment ?
Cites Auro · 2016-08 · Stack Overflow · on “Java Maven project won't build & get deployed to Heroku”
I'm quite new to both Heroku & Java EE projects. I'm trying to test a simple RESTful web app in Java & deploy it on Heroku. It's creating an app, but it's not building it, neither is it deploying it. I suspect there's something I am missing. I looked around a bit in StackOverflow, the couldn't figure out anything from the posts. I created a simple Java Maven project following this [this video] tutorial here on YouTube! Then I tried following this [LINK] tutorial here for deployment on my app & this is where things went wrong. Running the pom.xml as Maven Build is throwing the following error in the console [INFO] Scanning for projects... [INFO] Downloading: [LINK] [INFO] Downloading: [LINK] [INFO
Cites kenwood · 2017-01 · Stack Overflow · on “Created a maven project with jersey archtype but fail to build due to missing jersery servlet library”
I m trying to create a maven project with the jersey archtype by following the tutorial below. [this video] The maven project fails to build with a number of errors. The root cause, I believe, is this one: The container 'Maven Dependencies' references non existing library 'C:\Users\XXXXXXXX.m2\repository\com\sun\jersey\jersey-servlet\1.20-SNAPSHOT\jersey-servlet-1.20-SNAPSHOT.jar' The jar doesn't exist, in fact the \jersey... folder is missing. I am using eclipse, and have installed m2e. What else do I need to bring the jersey dependent files onto the computer?
Cites rehas · 2017-06 · Stack Overflow · on “Web application running without a name”
I just made a simple web application (using java and jersey as i want to start making a rest endpoint). I am working on intelliJ idea I used this tutorial [this video] to build it. I gave the project a name, group id and artifact. Now when i run it in my local tomcat server, it runs successfully but does not show me the name of my web application in URL. It only shows localhost:8080 as can be seen in the screen shot below. [LINK] Why is it displayed like that? How can i give my application a name/URL?
Cites JGFMK · 2017-11 · Stack Overflow · on “JAX-RS (Jersey 2 implementation) content negotiation with URL extension .xml or .json”
I've seen a Java RESTFUL webservice, that allowed the content-type to be requested in the URL with an extension at the end, such as .xml .json This is the style of content negotiation I am striving to achieve in my own Web Service. I am aware of the @Produces annotation, and the fact a method can resolve multiple types with the (value = {}) syntax, by adding an Accept header, say with Postman, the Chrome extension. But I'm not sure how to effectively extract out the information in one method, and delegate to another method. I'm assuming REGEX's can be use with @Path and @PathParam , but my attempts to do this have yet to be fruitful. Can anyone provide an example? This is my attempt thus far
Cites LosmiNCL · 2020-10 · Stack Overflow · on “Tomcat 404 error when hyperlink is clicked (using Jersey)”
I was watching this video tutorial for building RESTful web app: [this video] I've done everything like in video, so I'm in same situation. In short, I've created Maven project in Eclipse, with jersey-quickstart-webapp archtype. I'm using Tomcat 9.0 and I've added my project in it. Just like in the video, I've got welcoming Jersey page. But when I click in link, instead of "Got it" message I get 404 Not Found. Type: Status report Description: The origin server did not find a current representation for the target resource or is not willing to disclose that one exists. I haven't write any code, just like in the video. I think that Tomcat can't find webapi/myresource file This is my MyResource java