How to implement jaxrs Application without web.xml

java, jax-rs, jersey, rest

Solution

You need to deploy your application into a Servlet 3.0 compliant container to take advantage of this functionality. Try GlassFish 3.x or Tomcat 7.

Problem

I am trying to deploy a really simple jaxrs application with without a web.xml config and cannot get it working. My URL I'd expect to access is serverandport/{appname}/rest/welcomes/hello and I think I must be missing something dead obvious. Application ``` @ApplicationPath("/rest") public class EngineApp extends Application { @Override public Set<Class<?>> getClasses() { Set<Class<?>> s = new HashSet<Class<?>>(); s.add(RestTestImpl.class); return s; } } ``` Resource ``` @Path("/welcomes") public class RestTestImpl { @GET @Path("hello") public String sayPlainHello() { return "Hi from Rest"; } } ``` POM snippet ``` <groupId>org.apache.maven.plugins</groupId> <artifactId>maven-war-plugin</artifactId> <version>2.1.1</version> <configuration> <failOnMissingWebXml>false</failOnMissingWebXml> </configuration> ``` Edit: Further to the response below, I tried with an empty web.xml and also with the following web.xml. Bother also return 404, however the xml below states "Servlet javax.ws.rs.core.Application is not available": web.xml ``` <?xml version="1.0" encoding="UTF-8"?> <web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" version="3.0" 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_2_5.xsd" > <servlet> <servlet-name>javax.ws.rs.core.Application</servlet-name> <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping> <servlet-name>javax.ws.rs.core.Application</servlet-name> <url-pattern>/*</url-pattern> </servlet-mapping> </web-app> ```

Original source

Related problems