Question

我正试图联系我的控制器：

package com.atmWebApp.controllers;

import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.servlet.ModelAndView;

@Controller
public class FirstController {

        @RequestMapping("/welcome/")
        public ModelAndView helloWorld() {

            String message = "<br><div align='center'>" + "<h1>Hello World, Spring 3.2.1 Example by Crunchify.com<h1> <br>";
            message += "<a href='http://crunchify.com/category/java-web-development-tutorial/'>More Examples</a>";
            return new ModelAndView("welcome", "message", message);
        }

        @RequestMapping("/")
        public ModelAndView helloWorld1() {

            String message = "<br><div align='center'>" + "<h1>Hello World, Spring 3.2.1 Example by Crunchify.com<h1> <br>";
            message += "<a href='http://crunchify.com/category/java-web-development-tutorial/'>More Examples</a>";
            return new ModelAndView("welcome", "message", message);
        }
}

APP-servlet.xml中：

<?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:context="http://www.springframework.org/schema/context"
    xsi:schemaLocation="http://www.springframework.org/schema/beans

http://www.springframework.org/schema/beans/spring-beans-3.0.xsd

http://www.springframework.org/schema/context 

http://www.springframework.org/schema/context/spring-context-3.0.xsd">

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

    <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/jsp/" />
        <property name="suffix" value=".jsp" />
    </bean>
</beans>

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>AtmWebApp</display-name>
    <welcome-file-list>
        <welcome-file>index.jsp</welcome-file>
    </welcome-file-list>

    <servlet>
        <servlet-name>app</servlet-name>
        <servlet-class>
            org.springframework.web.servlet.DispatcherServlet
        </servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>app</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>
</web-app><?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>AtmWebApp</display-name>
    <welcome-file-list>
        <welcome-file>index.jsp</welcome-file>
    </welcome-file-list>

    <servlet>
        <servlet-name>app</servlet-name>
        <servlet-class>
            org.springframework.web.servlet.DispatcherServlet
        </servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>app</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>
</web-app>

具体来说，我想启动我的服务器并将我的root设置为localhost：8080 /，但我甚至无法点击FirstController。所以我的主要问题是：为什么我没有在FirstController中使用任何一种方法？

我认为我的问题出在app-server.xml或web.xml文件中。我查看了堆栈溢出here，here，here和here，但我不知道我的做法与这些人不同。

任何帮助？

Answer 1

据我所知，您还应该在控制器上设置@RequestMapping(..)。因此你应该有这样的事情：

@Controller
@RequestMapping("/")
public class FirstController {
    ...

同样，您还可以检查this网址映射的响应。

Answer 2

在web.xml文件中，在servlet映射下尝试将url-Pattern从/更改为/*.

Spring MVC调度程序servlet，没有应用程序名称的URI

2 个答案: