找不到url Observer的请求

时间:2011-12-16 09:43:33

标签: spring rest maven cxf

我正在使用Apache CXF的示例REST服务,但不知怎的,我无法调用该服务。我的实现类是,

package com.ananth.lab.cfxrest.service;

import com.ananth.lab.cfxrest.vo.Address;
import com.ananth.lab.cfxrest.vo.Employee;
import org.springframework.stereotype.Service;

import javax.jws.WebService;
import javax.ws.rs.*;
import java.util.ArrayList;
import java.util.Collection;
import java.util.List;


@Path("/cservice")
@Produces("application/xml")
public class EmployeeService {

    @GET
    @Path("/emp")

    public Employee getEmployee() {

        Address address1 = new Address();
        address1.setCity("Chennai");
        address1.setZip(63);
        List<Address> list = new ArrayList<Address>();
        Address address2 = new Address();
        address2.setCity("Bangalore");
        address2.setZip(49);
        list.add(address1);
        list.add(address2);
        Employee emp = new Employee();
        emp.setAddress(list);
        emp.setEmployeeId("001");
        emp.setEmployeeName("Ananth");

        return emp;
    }

}

我的web.xml文件是,

<!DOCTYPE web-app PUBLIC
 "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
 "http://java.sun.com/dtd/web-app_2_3.dtd" >

<web-app>
  <display-name>Hello world REST service with apache cxf</display-name>
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>WEB-INF/beans.xml,WEB-INF/applicationContext.xml</param-value>
    </context-param>

    <listener>
        <listener-class>
            org.springframework.web.context.ContextLoaderListener
        </listener-class>
    </listener>
    <servlet>
        <servlet-name>CXFServlet</servlet-name>
        <display-name>CXF Servlet</display-name>
        <servlet-class>
            org.apache.cxf.transport.servlet.CXFServlet
        </servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>CXFServlet</servlet-name>
        <url-pattern>/*</url-pattern>
    </servlet-mapping>



</web-app>

我部署在Tomcat中,上下文路径是&#34; Lab&#34;。所以我试图访问像

这样的服务
http://localhost:8080/Lab/cservice/emp

我正在

No service was found.

2 个答案:

答案 0 :(得分:1)

我认为你应该确保你的Employee类有正确的“@”评论。如下所示:

@XmlRootElement(name="cservice")

public class Employee (){
    private String employeeId;
    private String employeeName;
    ...


@XmlElement(name="employeeId")
    public void setEmployeeId(String employeeId){
        ...
    }
    public String getEmployeeId(){
        return this.employeeId;
    }

@XmlElement(name="employeeName")
    public void setEmployeeName(String employeeId){
        ...
    }
    public String getEmployeeName(){
        return this.employeeName
    }
...
...
}

我建议您查看WEB-INF / beans.xml&amp; WEB-INF / applicationContext.xml中。

答案 1 :(得分:-1)

确保您的请求网址端点与您的服务器端点匹配。

将webl和beans.xml中的url结合起来产生终结点