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스프링 -WebMVC 및 아파치 CXF를 사용하여 프로젝트를 설정하려고하지만 항상 웹 서비스를 호출하려고하면 "서비스를 찾을 수 없습니다."라는 메시지가 나타납니다.Spring WebMVC와 Apache CXF를 결합하려고하지만 서비스를 찾을 수 없습니다. 왜?
의 web.xml :
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 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>RESTful service</display-name> <description></description> <session-config> <session-timeout>30</session-timeout> </session-config> <context-param> <param-name>webAppRootKey</param-name> <param-value>cxf.rest.root</param-value> </context-param> <context-param> <param-name>log4jConfigLocation</param-name> <param-value>/WEB-INF/classes/log4j.properties</param-value> </context-param> <listener> <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class> </listener> <context-param> <param-name>contextConfigLocation</param-name> <param-value>/WEB-INF/appContext*.xml</param-value> </context-param> <servlet> <servlet-name>CXFServlet</servlet-name> <servlet-class>org.apache.cxf.transport.servlet.CXFServlet</servlet-class> <load-on-startup>1</load-on-startup> </servlet> <servlet> <servlet-name>mvc-dispatcher</servlet-name> <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> <load-on-startup>2</load-on-startup> </servlet> <servlet-mapping> <servlet-name>mvc-dispatcher</servlet-name> <url-pattern>/</url-pattern> </servlet-mapping> <servlet-mapping> <servlet-name>CXFServlet</servlet-name> <url-pattern>/rs/*</url-pattern> </servlet-mapping> </web-app>MVC-디스패처 여기
org.apache.cxf.transport.servlet.ServletController invoke WARNING: Can't find the request for http://localhost:8080/HelloWebService/rs/user-service/users's Observer프로젝트 구성은 다음과 같습니다 아파치 톰캣 로그에서
는 경고가 -servlet.xml appContext.xml이<?xml version="1.0" encoding="UTF-8"?> <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:jaxrs="http://cxf.apache.org/jaxrs" xsi:schemaLocation=" http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd http://cxf.apache.org/jaxrs http://cxf.apache.org/schemas/jaxrs.xsd" default-lazy-init="false"> <import resource="classpath:META-INF/cxf/cxf.xml"/> <import resource="classpath:META-INF/cxf/cxf-servlet.xml"/> <jaxrs:server id="userService" address="/"> <jaxrs:serviceBeans> <ref bean="userService" /> </jaxrs:serviceBeans> <jaxrs:extensionMappings> <entry key="xml" value="application/xml" /> <entry key="json" value="application/json" /> </jaxrs:extensionMappings> </jaxrs:server> <bean id="userService" class="org.home.playground.services.UserService"/> </beans>
내가 웹 애플리케이션의 첫 번째 페이지로이 index.jsp를 사용하고 0
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:mvc="http://www.springframework.org/schema/mvc"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="
http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans.xsd
http://www.springframework.org/schema/mvc
http://www.springframework.org/schema/mvc/spring-mvc.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context.xsd">
<context:component-scan base-package="org.home.playground" />
<mvc:annotation-driven/>
<bean id="viewResolver"
class="org.springframework.web.servlet.view.UrlBasedViewResolver">
<property name="viewClass"
value="org.springframework.web.servlet.view.JstlView" />
<property name="prefix" value="/WEB-INF/views/" />
<property name="suffix" value=".jsp" />
</bean>
. 평온한 웹 서비스처럼 보이는
<html>
<body>
<h2>CXF RESTful services and Spring-WebMVC Test page</h2>
<a href="rs/user-service/users">get all users</a>
<br />
</body>
</html>
: :이 index.jsp에서 다음 코드를 포함
package org.home.playground.interfaces;
import java.io.IOException;
import javax.servlet.http.HttpServletResponse;
import javax.ws.rs.Consumes;
import javax.ws.rs.FormParam;
import javax.ws.rs.GET;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.PathParam;
import javax.ws.rs.Produces;
import javax.ws.rs.core.Context;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.Response;
import org.home.playground.apputils.UserCollection;
import org.home.playground.models.User;
@Path("/user-service/")
@Produces("application/xml")
public interface IUserService {
@GET
@Path("/users")
@Produces({"application/xml", "application/json"})
public UserCollection getUsers();
@GET
@Path("/user/{id}")
public User getUser(@PathParam("id") Integer id);
@GET
@Path("https://stackoverflow.com/users/bad")
public Response getBadRequest();
@POST
@Path("/new")
@Produces(MediaType.TEXT_HTML)
@Consumes(MediaType.APPLICATION_FORM_URLENCODED)
public void newUser(
@FormParam("id") Integer id,
@FormParam("name") String name,
@Context HttpServletResponse servletResponse
) throws IOException;
}
내 스프링 webMVC 컨트롤러가 다음과 같습니다
package org.home.playground.web.controllers;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.servlet.ModelAndView;
@Controller
public class HelloWorldController {
@RequestMapping("/")
protected ModelAndView showHelloWorldView(HttpServletRequest request, HttpServletResponse response) throws Exception {
ModelAndView model = new ModelAndView("index");
return model;
}
@RequestMapping("/postUser")
protected ModelAndView postUserForm(HttpServletRequest request, HttpServletResponse response) throws Exception {
ModelAndView model = new ModelAndView("postUser");
return model;
}
}
당신은 이유를 알고 있습니까 웹 서비스를 호출 할 수 없습니까? Spring-WebMVC와 Apache CXF를 결합 할 수 있습니까? 아니면 이러한 프레임 워크가 하나의 프로젝트에서 호환되지 않습니까? 좀 더 자세한 정보가 필요하면 알려 주시면 제가 제공 할 것입니다. 귀하의 도움에 감사 드리며 미리 감사드립니다.
새로운 응용 프로그램입니까? 왜 오래된 XML 설정으로 귀찮아, 당신은 XML 무료 구성을 만들 수 있습니다. –
나는 옛날 학교이고 나는 XML로 나의 구성을 좋아한다. 그럼에도 불구하고 이것은 내 주요 문제의 주제가 아닙니다 ... ;-) – F4k3d