Question

控制器无法通过网址“ / hello”解析-错误404？

错误在哪里？

С您显示了一个有效的示例吗？

描述原始服务器找不到目标资源的当前表示，或者不愿意透露该资源的存在。

Web xml设置

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.1" 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">
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/applicationContext.xml</param-value>
    </context-param>
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>
    <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>2</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>

    <session-config>
        <session-timeout>
            30
        </session-timeout>
    </session-config>
    <welcome-file-list>
        <welcome-file>redirect.jsp</welcome-file>
    </welcome-file-list>
</web-app>

servlet

dispatcher-servlet

<?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:p="http://www.springframework.org/schema/p"
xmlns:aop="http://www.springframework.org/schema/aop"
xmlns:tx="http://www.springframework.org/schema/tx"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.5.xsd
 http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-2.5.xsd
 http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-2.5.xsd
 http://www.springframework.org/schema/jdbc http://www.springframework.org/schema/jdbc/spring-jdbc-4.0.xsd
 http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-4.0.xsd
 http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-4.0.xsd">


    <bean class="org.springframework.web.servlet.mvc.support.ControllerClassNameHandlerMapping"/>


    <!--
    Most controllers will use the ControllerClassNameHandlerMapping above, but
    for the index controller we are using ParameterizableViewController, so we must
    define an explicit mapping for it.
    -->
    <bean id="urlMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
        <property name="mappings">
            <props>
                <prop key="index.htm">indexController</prop>
            </props>
        </property>
    </bean>

    <bean id="viewResolver"
          class="org.springframework.web.servlet.view.InternalResourceViewResolver"
          p:prefix="/WEB-INF/jsp/"
          p:suffix=".jsp" />

    <!--
    The index controller.
    -->
    <bean name="indexController"
          class="org.springframework.web.servlet.mvc.ParameterizableViewController"
          p:viewName="index" />

    <context:component-scan base-package="controller"/>

</beans>

控制器

控制器

package controller;

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */


import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;

/**
 *
 * @author nn
 */
@Controller
public class TestController {

    @RequestMapping(value = "/hello", method = RequestMethod.GET)
    public String add(){
        System.out.println("text");
        System.exit(0);
        return "hello";
    }
}

hello.jsp

<%@page contentType="text/html" pageEncoding="UTF-8"%>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
    "http://www.w3.org/TR/html4/loose.dtd">

<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <title>Welcome to Spring Web MVC project</title>
    </head>

    <body>
        <p>Hello! This is the default welcome page for a Spring Web MVC project.</p>
        <p><i>To display a different welcome page for this project, modify</i>
            <tt>index.jsp</tt> <i>, or create your own welcome page then change
                the redirection in</i> <tt>redirect.jsp</tt> <i>to point to the new
                welcome page and also update the welcome-file setting in</i>
            <tt>web.xml</tt>.</p>
    </body>
</html>

通过url / hello获得的代码和启动控制器的设置应该是什么

Spring Controller无法通过URL解析

0 个答案: