CODEWITHSUNDEEP
javaweb-servicestomcatjersey
Rajeswari Kotikalapudi
Rajeswari Kotikalapudi

Reputation: 592

Restful WebServices in Java using Eclipse, Tomcat and Jersey

How to create simple webserver in Java using Eclipse, Tomcat and Jersey i.e steps to follow?

We are creating simple webserver using the below links:

but we got an error like this: 

java.lang.ClassNotFoundException: com.sun.jersey.spi.container.servlet.ServletContainer

Upvotes: 1

Views: 6661

Answers (2)

Robin Wieruch
Robin Wieruch

Reputation: 15908

Copy all your Jersey jars, including jersey-servlet-1.12.jar, in your lib folder. Look that you have included it in the build path.

Upvotes: 0

anvarik
anvarik

Reputation: 6487

Have maven running. Then run this command(press enter if it asks sth):

mvn archetype:generate -DgroupId=com.test.rest -DartifactId=test -DarchetypeArtifactId=maven-archetype-webapp

It will create you a simple webapp. Now create the source package as src/main/java/com/test/rest, and create a simple class as following with a name "test" in it:

 package com.test.rest;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.PathParam;
import javax.ws.rs.core.Response;

@Path("/test")
public class test{

@GET
@Path("/{param}")
public Response getMsg(@PathParam("param") String msg) {

    String output = "Jersey say : " + msg;

    return Response.status(200).entity(output).build();

    }

}

At that point you should get errors, resolve them by adding this dependency to your pom:

    <dependency>
        <groupId>com.sun.jersey</groupId>
        <artifactId>jersey-server</artifactId>
        <version>1.8</version>
    </dependency>

you can run a dummy "mvn clean install" so that maven will download the repository and your errors will disappear.

Now, go to webapp/WEB-INF and configure your web.xml as follows:

<web-app id="WebApp_ID" version="2.4"
xmlns="http://java.sun.com/xml/ns/j2ee" 
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee 
http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>Restful Web Application</display-name>

<servlet>
    <servlet-name>jersey-serlvet</servlet-name>
    <servlet-class>
                 com.sun.jersey.spi.container.servlet.ServletContainer
            </servlet-class>
    <init-param>
         <param-name>com.sun.jersey.config.property.packages</param-name>
         <param-value>com.test.rest</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>jersey-serlvet</servlet-name>
    <url-pattern>/rest/*</url-pattern>
</servlet-mapping>

here we said which classes to be loaded and also gave a small prefix with "/rest". so your webservice will start with this prefix.

Now you are ready, build the app, and add the jar file under tomcat/webapps folder. when you run your tomcat you can reach to your webservice via:

(url_to_tomcat_server/jar_name/prefix_at_web_xml/prefix_at_java_rest_class/dummy_text_requested_byclass)

localhost:8080/test/rest/test/blabla

Note: tested and running

Upvotes: 3

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