CODEWITHSUNDEEP
javaweb-servicesrestmaventomcat
user818700
user818700

Reputation:

REST Webservice 404 on Tomcat 8 using maven

I'm getting a 404 NOT FOUND exception when I try and run my webservice on tomcat 8. My pom looks like this:

<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
    <modelVersion>4.0.0</modelVersion>
    <groupId>com.deangrobler.testrest</groupId>
    <artifactId>TestRest</artifactId>
    <version>0.0.1-SNAPSHOT</version>
    <dependencies>
        <dependency>
            <groupId>javax.ws.rs</groupId>
            <artifactId>javax.ws.rs-api</artifactId>
            <version>2.0</version>
        </dependency>
        <dependency>
            <groupId>org.glassfish.jersey.containers</groupId>
            <artifactId>jersey-container-servlet</artifactId>
            <version>2.10.1</version>
        </dependency>
    </dependencies>
</project>

And my web.xml looks like:

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns="http://java.sun.com/xml/ns/javaee"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
    id="WebApp_ID" version="3.0">
    <display-name>TestRest</display-name>

    <servlet>
        <servlet-name>jersey-helloworld-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.javacodegeeks.enterprise.rest.jersey</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

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

</web-app>

And my annotated class:

package com.deangrobler.testrest;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;

@Path("/hellorest")
public class HelloRest {

    @GET
    @Produces("text/plain")
    public String hello () {
        return "Hello there developer!";
    }

}

I expect to then being able to hit it at

http://localhost:8080/TestRest/rest/hellorest/

But... I just get that 404 error.. Something I'm missing? I'd appreciate any help :)

Upvotes: 0

Views: 1374

Answers (3)

K R Anushree
K R Anushree

Reputation: 21

You seem to be using Jersey 2.0 version. So in your web.xml you need to change the servlet part.

<servlet>
    <servlet-name>JSON RESTful Service</servlet-name>
    <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
    <init-param>
        <param-name>jersey.config.server.provider.packages</param-name>
        <param-value>com.samplerest.java</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

Remember to change the param-value to the name of the package where you have your java code.

This will help :- 404 error with Jersey REST Service on Tomcat

Upvotes: 0

Jens
Jens

Reputation: 69440

I think the name of your war is TestRest-0.0.1-SNAPSHOT. So you have to use the url: http://localhost:8080/TestRest-0.0.1-SNAPSHOT/rest/hellorest/

Upvotes: 0

mkrakhin
mkrakhin

Reputation: 3486

Param com.sun.jersey.config.property.packages is meant to denote packages containing your REST classes. So param-value should be com.deangrobler.testrest.
Check documentation for ServletContainer.

Upvotes: 1

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