Question

我正在尝试运行spring应用程序，但是我收到了404错误。有人可以帮忙吗？我从mkyong网站上获取此代码。他给出的例子是工作，但我试图从头开始做任何事都不起作用。

package com.mkyong.common.controller;
import org.springframework.stereotype.Controller;    
import org.springframework.ui.ModelMap;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;

@Controller
@RequestMapping("/welcome")
public class HelloController {

    @RequestMapping(method = RequestMethod.GET)
    public String printWelcome(ModelMap model) {

        model.addAttribute("message", "Spring 3 MVC Hello World");
        return "hello";

    }

}

hello.jsp
<html>
<body>
    <h1>Message : ${message}</h1>   
</body>
</html>

mvc-dispatcher-servlet.xml
<beans xmlns="http://www.springframework.org/schema/beans"
    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-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.mkyong.common.controller" />

    <bean
        class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix">
            <value>/WEB-INF/pages/</value>
        </property>
        <property name="suffix">
            <value>.jsp</value>
        </property>
    </bean>

</beans>

web.xml
<web-app id="WebApp_ID" version="2.4"
    xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee 
    http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">

    <display-name>Spring Web MVC Application</display-name>

    <servlet>
        <servlet-name>mvc-dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>mvc-dispatcher</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/mvc-dispatcher-servlet.xml</param-value>
    </context-param>

    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <welcome-file-list>
        <welcome-file>index.jsp</welcome-file>
    </welcome-file-list>

</web-app>

Answer 1

在web.xml中添加*到servlet的URL模式：

<servlet-mapping>
    <servlet-name>mvc-dispatcher</servlet-name>
    <url-pattern>/*</url-pattern>
</servlet-mapping>

Answer 2

在您的方法上使用@RequestMapping（＆＃34; / welcome＆＃34;）而不是类

Spring 404错误

2 个答案: