REST with Could not find error

Question

I am trying to create a REST application, in Java web, very simple, using Tomcat 8 as a server. However, when trying to access the application:

http://localhost:8080/aplicacao/rest/teste

She brings me the following mistake:

Mar 03, 2015 9:18:39 AM org.jboss.resteasy.core.ExceptionHandler 
SEVERE: failed to execute
javax.ws.rs.NotFoundException: Could not find resource for full path: http://localhost:8080/aplicacao/rest/teste

According to the searches I did, my settings files seem to be correct:

Application class:

@Path("/teste")
public class TesteService {

    @GET
    @Produces({MediaType.APPLICATION_JSON, MediaType.APPLICATION_XML})
    public Response listar() {

        List<String> listaList = new ArrayList<String>();

        //...populo a lista

        return Response.ok(listaList).build();
    }
}

Web file.xml:

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd" version="3.1">
  <display-name>aplicacao</display-name>
  <welcome-file-list>
    <welcome-file>index.html</welcome-file>
    <welcome-file>index.htm</welcome-file>
    <welcome-file>index.jsp</welcome-file>
    <welcome-file>default.html</welcome-file>
    <welcome-file>default.htm</welcome-file>
    <welcome-file>default.jsp</welcome-file>
  </welcome-file-list>
  <display-name>aplicacao rest</display-name>
     <context-param>
        <param-name>resteasy.servlet.mapping.prefix</param-name>
        <param-value>/rest</param-value>
    </context-param>
    <context-param>
        <param-name>resteasy.scan</param-name>
        <param-value>true</param-value>
    </context-param>

    <servlet>
        <servlet-name>RestServlet</servlet-name>
        <servlet-class>org.jboss.resteasy.plugins.server.servlet.HttpServletDispatcher</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>RestServlet</servlet-name>
        <url-pattern>/rest/*</url-pattern>
    </servlet-mapping>
</web-app>

pom.xml file:

<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>aplicacao</groupId>
    <artifactId>aplicacao</artifactId>
    <version>0.0.1-SNAPSHOT</version>
    <packaging>war</packaging>
    <build>
        <sourceDirectory>src</sourceDirectory>
        <plugins>
            <plugin>
                <artifactId>maven-compiler-plugin</artifactId>
                <version>3.1</version>
                <configuration>
                    <source>1.7</source>
                    <target>1.7</target>
                </configuration>
            </plugin>
            <plugin>
                <artifactId>maven-war-plugin</artifactId>
                <version>2.4</version>
                <configuration>
                    <warSourceDirectory>WebContent</warSourceDirectory>
                    <failOnMissingWebXml>false</failOnMissingWebXml>
                </configuration>
            </plugin>
        </plugins>
    </build>
    <dependencies>
        <dependency>
            <groupId>org.jboss.resteasy</groupId>
            <artifactId>jaxrs-api</artifactId>
            <version>3.0.6.Final</version>
        </dependency>
        <dependency>
            <groupId>org.jboss.resteasy</groupId>
            <artifactId>resteasy-jaxrs</artifactId>
            <version>3.0.10.Final</version>
        </dependency>

    </dependencies>
</project>

Answer 1

Write down your method with @path("/"). This means that this is the root path of your resource. Example:

@Path("/teste")
public class TesteService {

    @Path("/")
    @GET
    @Produces({MediaType.APPLICATION_JSON, MediaType.APPLICATION_XML})
    public Response listar() {

        List<String> listaList = new ArrayList<String>();

        //...populo a lista

        return Response.ok(listaList).build();
    }
}