初学者Java REST 404错误

Question

package sampproj;

import javax.ws.rs.Consumes;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.GET;
import javax.ws.rs.Produces;
import javax.ws.rs.PathParam;

@Path("sampproj")
public class WSPostSer {
    public WSPostSer() {
        super();
    }

    @POST
    @Consumes("application/json")
    @Path("devpost")
    public String createTemp (String devName){
        System.out.println (" in reate");
        return "heloo" + devName;
    }

    @GET
    @Produces("text/plain")
    @Path("/uservalue")
    public String welcomeResource (@PathParam("uservalue") String username){    
        System.out.println("  welcomeResource ");   
        return "heloo Get " + username; 
    }
}

<?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>SampleRS</display-name>
   <servlet>
      <servlet-name>Jersey RESTful Application</servlet-name>
      <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
         <init-param>
            <param-name>jersey.config.server.provider.packages</param-name>
            <param-value>sampproj</param-value>
         </init-param>
      </servlet>
   <servlet-mapping>
   <servlet-name>Jersey RESTful Application</servlet-name>
      <url-pattern>/rest/*</url-pattern>
   </servlet-mapping>  
</web-app>

在此输入图像说明

Answer 1

将方法更改为：

@GET
@Produces("text/plain")
@Path("{uservalue}")
public String welcomeResource (@PathParam("uservalue") String username){
    System.out.println("  welcomeResource ");
    return "heloo Get " + username;
}

war文件名为wswebapp.war，因此在浏览器中键入以下URL：

http://localhost:8081/WSwebapp/rest/sampproj/me

您应该看到heloo Get me:)