Question

尽管看了好几个小时，我仍无法弄清为什么会出现404错误。

HTTP Status 404 – Not Found
Type Status Report

Message /

Description The origin server did not find a current representation for the target resource or is not willing to disclose that one exists.

Apache Tomcat/9.0.24

我有一个非常简单的spring MVC应用程序，但我真的看不到错误是什么。请帮助我失去理智

结构：

SpringMVCDemo

src
- 主要
  - java
    - com.springmvc.demo
      - HomeController.java
- WEB-INF
  - 视图
    - main-menu.jsp
- spring-mvc-demo-servlet.xml
- web.xml

我的控制器： HomeController.java

package com.springmvc.demo;

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

@Controller
public class HomeController {

@RequestMapping("/")
public String showMyPage(){
    return "main-menu"; // view name
}

我的观点： main-menu.jsp

<!DOCTYPE html>
<html>
<body>

<h2> Spring MVC Demo - main-menu</h2>

</body>
</html>

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://xmlns.jcp.org/xml/ns/javaee"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
id="WebApp_ID" version="3.1">

<display-name>spring-mvc-demo</display-name>

<absolute-ordering />

<!-- Spring MVC Configs -->

<!-- Step 1: Configure Spring MVC Dispatcher Servlet -->
<servlet>
    <servlet-name>dispatcher</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>WEB-INF/spring-mvc-demo-servlet.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

<!-- Step 2: Set up URL mapping for Spring MVC Dispatcher Servlet -->
<servlet-mapping>
    <servlet-name>dispatcher</servlet-name>
    <url-pattern>/</url-pattern>
</servlet-mapping>

</web-app>

spring-mvc-demo-servlet.xml

<?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:context="http://www.springframework.org/schema/context"
xmlns:mvc="http://www.springframework.org/schema/mvc"
xsi:schemaLocation="
    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.xsd
    http://www.springframework.org/schema/mvc
    http://www.springframework.org/schema/mvc/spring-mvc.xsd">

<!-- Step 3: Add support for component scanning -->
<context:component-scan base-package="com.springmvc.demo" />

<!-- Step 4: Add support for conversion, formatting and validation support -->
<mvc:annotation-driven/>

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

</beans>

Answer 1

在这种情况下，我的问题通过重新创建从模块爆炸的构件SpringMVCdemo：war得以解决。

我一开始并不完全了解错误，但希望以后能对某人有所帮助

Spring MVC，为什么会出现404错误？

1 个答案: