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友好的网址映射问题-Java Spring

2019-10-09 19:22:29 作者：互联网

我在web.xml上的所有页面都都显示为404的错误中挣扎,可能存在根路径,但是我不确定其设置位置等.

这是我当前的web.xml

<?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" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
  <display-name>Spring3MVC</display-name>
  <welcome-file-list>
    <welcome-file>index.jsp</welcome-file>
  </welcome-file-list>
  <servlet>
    <servlet-name>spring</servlet-name>
    <servlet-class>
                  org.springframework.web.servlet.DispatcherServlet
              </servlet-class>
    <load-on-startup>1</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>spring</servlet-name>
    <url-pattern>/*</url-pattern>

  </servlet-mapping>
</web-app>

我的监听器控制器是这样的

/*
 * User
*/
@RequestMapping(value={"/user/{id}"}, method=RequestMethod.GET)
public ModelAndView profileDisplay(
        HttpServletRequest request, 
        HttpServletResponse response,
        @RequestParam(value="id", required=false) String id
) throws UnknownHostException, MongoException {
    ServiceSerlvet.appendSesssion(request);
    //get search ALL users
    BasicDBObject searchQuery = new BasicDBObject();
        searchQuery.put("_id", new ObjectId(id));
    List<DBObject> searchResponse = PersonController.searchUsers(searchQuery);      

    //System.out.println("response from search user method: "+searchResponse);

        return new ModelAndView("user", "people", searchResponse);
}

这是即将出现的当前错误.它为什么没有映射,我该如何解决呢？

INFO: Server startup in 5904 ms
01-Nov-2012 19:40:21 org.springframework.web.servlet.DispatcherServlet noHandlerFound
WARNING: No mapping found for HTTP request with URI [/springApp21] in DispatcherServlet with name 'spring'
01-Nov-2012 19:40:22 org.springframework.web.servlet.mvc.support.DefaultHandlerExceptionResolver handleNoSuchRequestHandlingMethod
WARNING: No matching handler method found for servlet request: path '/user', method 'GET', parameters map['id' -> array<String>['4fa6eddc0234964172522248']]
01-Nov-2012 19:40:24 org.springframework.web.servlet.mvc.support.DefaultHandlerExceptionResolver handleNoSuchRequestHandlingMethod
WARNING: No matching handler method found for servlet request: path '/user', method 'GET', parameters map['id' -> array<String>['4fa6eddc0234964172522248']]

解决方法:

在回答我的问题之前,我曾不屑一顾.我现在可以访问我的其中一个春季应用程序.这是一个更好的配置.

注意对web.xml的更改,我很抱歉,但是映射到/ *会使调度程序解决您的所有请求.从某种意义上说,您创建了一个循环,您的初始映射将由调度程序转发到控制器,然后控制器将使用视图解析器来映射应转发请求的位置.映射到/ *导致视图解析器映射由调度程序处理.

更改为/会导致所有未映射的url由调度程序处理,因此您的初始映射由调度程序处理,调度程序将其发送到控制器,并且您的viewresolver创建的映射将映射到.jsp,导致无法选择它由调度员处理.我很抱歉.

在web.xml

<?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" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
  <display-name>Spring3MVC</display-name>
  <welcome-file-list>
    <welcome-file>index.jsp</welcome-file>
  </welcome-file-list>
  <servlet>
    <servlet-name>spring</servlet-name>
    <servlet-class>
                  org.springframework.web.servlet.DispatcherServlet
              </servlet-class>
    <load-on-startup>1</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>spring</servlet-name>
    <url-pattern>/</url-pattern>

  </servlet-mapping>
</web-app>

spring-config.xml(必须更改组件扫描)

<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
    xmlns:beans="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:context="http://www.springframework.org/schema/context"
    xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.1.xsd
        http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
        http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.1.xsd">

    <!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure -->

    <!-- Enables the Spring MVC @Controller programming model -->   
    <annotation-driven/>

        <!-- Handles HTTP GET requests for /resources/** by efficiently serving up static resources in the ${webappRoot}/resources directory -->
    <resources location="/resources/" mapping="/resources/**"/> 

    <!-- Resolves views selected for rendering by @Controllers to .jsp resources in the /WEB-INF/views directory -->
    <beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <beans:property name="prefix" value="/WEB-INF/views/"/>
        <beans:property name="suffix" value=".jsp"/>
    </beans:bean>

    <context:component-scan base-package="package.with.controllers" />

</beans:beans>

调节器

@RequestMapping(value={"/user/{id}"}, method=RequestMethod.GET)
public ModelAndView profileDisplay(
        HttpServletRequest request, 
        HttpServletResponse response,
        @RequestParam(value="id", required=false) String id
) throws UnknownHostException, MongoException {
    ServiceSerlvet.appendSesssion(request);
    //get search ALL users
    BasicDBObject searchQuery = new BasicDBObject();
        searchQuery.put("_id", new ObjectId(id));
    List<DBObject> searchResponse = PersonController.searchUsers(searchQuery);      

    //System.out.println("response from search user method: "+searchResponse);

        //This should display "WEB-INF/views/user.jsp" you may need to adjust.
        return new ModelAndView("user", "people", searchResponse);
}

标签：friendly-url,java,spring,spring-mvc,mapping
来源： https://codeday.me/bug/20191009/1881438.html