我的Spring Web应用程序有问题:它显示Apache Tomcat / 4.0.6-HTTP状态404-/ spring-mvc-example /((请求的资源(/ spring-mvc-example /)不可用)。 )我为这个错误而疯狂,我不知道该怎么办。 我正在使用STS。 谢谢您的帮助! 这是文件夹的组织方式。 https://i.stack.imgur.com/ded48.png
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_4_0.xsd" id="WebApp_ID"
version="4.0">
<display-name>spring-mvc-example</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<display-name>spring-mvc-example</display-name>
<!-- Add Spring MVC DispatcherServlet as front controller -->
<servlet>
<servlet-name>spring</servlet-name>
<servlet-class>
org.springframework.web.servlet.DispatcherServlet
</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring-servlet.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>spring</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
spring-servlet.xml
<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:beans="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/mvc
http://www.springframework.org/schema/mvc/spring-mvc.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.xsd">
<!-- DispatcherServlet Context: defines this servlet's request-processing
infrastructure -->
<!-- Enables the Spring MVC @Controller programming model -->
<annotation-driven />
<context:component-scan base-package="com.spring" />
<!-- 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>
pom.xml
<project xmlns="http://maven.apache.org/POM/4.0.0"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0
http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<groupId>spring-mvc-example</groupId>
<artifactId>spring-mvc-example</artifactId>
<version>0.0.1-SNAPSHOT</version>
<packaging>war</packaging>
<name>Spring MVC Example</name>
<description>Picco esempio</description>
<!-- Add Spring Web and MVC dependencies -->
<dependencies>
<dependency>
<groupId>org.springframework</groupId>
<artifactId>spring-webmvc</artifactId>
<version>4.3.9.RELEASE</version>
</dependency>
<dependency>
<groupId>org.springframework</groupId>
<artifactId>spring-web</artifactId>
<version>4.3.9.RELEASE</version>
</dependency>
<!-- Servlet -->
<dependency>
<groupId>javax.servlet</groupId>
<artifactId>servlet-api</artifactId>
<version>2.5</version>
<scope>provided</scope>
</dependency>
<dependency>
<groupId>javax.servlet.jsp</groupId>
<artifactId>jsp-api</artifactId>
<version>2.1</version>
<scope>provided</scope>
</dependency>
<dependency>
<groupId>javax.servlet</groupId>
<artifactId>jstl</artifactId>
<version>1.2</version>
</dependency>
</dependencies>
<build>
<plugins>
<plugin>
<artifactId>maven-compiler-plugin</artifactId>
<version>3.8.0</version>
<configuration>
<source>1.8</source>
<target>1.8</target>
</configuration>
</plugin>
<plugin>
<artifactId>maven-war-plugin</artifactId>
<version>3.2.1</version>
<configuration>
<warSourceDirectory>WebContent</warSourceDirectory>
</configuration>
</plugin>
</plugins>
HomeController.java
package com.spring.controller;
import java.text.DateFormat;
import java.util.Date;
import java.util.Locale;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.validation.annotation.Validated;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import com.spring.model.User;
@Controller
public class HomeController {
/**
* Simply selects the home view to render by returning its name.
*/
@RequestMapping(value = "/", method = RequestMethod.GET)
public String home(Locale locale, Model model) {
System.out.println("Home Page Requested, locale = " + locale);
Date date = new Date();
DateFormat dateFormat = DateFormat.getDateTimeInstance(DateFormat.LONG, DateFormat.LONG, locale);
String formattedDate = dateFormat.format(date);
model.addAttribute("serverTime", formattedDate);
return "home";
}
@RequestMapping(value = "/user", method = RequestMethod.POST)
public String user(@Validated User user, Model model) {
System.out.println("User Page Requested");
model.addAttribute("userName", user.getUserName());
return "user";
}
}
home.jsp
<%@ page language="java" contentType="text/html; charset=UTF-8"
pageEncoding="UTF-8"%>
<%@ taglib uri="http://java.sun.com/jsp/jstl/core" prefix="c"%>
<%@ page session="false"%>
<html>
<head>
<title>Home</title>
</head>
<body>
<h1>Hello world!</h1>
<P>The time on the server is ${serverTime}.</p>
<form action="user" method="post">
<input type="text" name="userName"><br>
<input type="submit" value="Login">
</form>
</body>
</html>
user.jsp
%@ page language="java" contentType="text/html; charset=UTF-8"
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>User Home Page</title>
</head>
<body>
<h3>Hi ${userName}</h3>
</body>
</html>
答案 0 :(得分:1)
两个问题:
<context:component-scan base-package="com.spring" />
更改为
<context:component-scan base-package="com.spring.*" />
<beans:property name="prefix" value="/WEB-INF/views/" />
更改为
<beans:property name="prefix" value="/WEB-INF/view/" />
(自
您的文件夹名为视图) 答案 1 :(得分:0)
将web.xml servlet映射更改为
<servlet-mapping>
<servlet-name>spring</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
然后在/WEB-INF/spring-servlet.xml
的contextConfigLocation文件中进行如下配置:
<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">
// to scan for annotation controllers, beans or configurations
<context:component-scan base-package="com.spring" />
<bean
class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix">
<value>/WEB-INF/view/</value>
</property>
<property name="suffix">
<value>.jsp</value>
</property>
</bean>
</beans>
您没有正确打包和部署应用。
我建议您将应用程序打包到适当的WAR文件中,将其放在/ webapps或/ WebContent文件夹中,然后启动Tomcat。
如果该软件包的名称为spring-mvc-example.war或[any-name] .war,则您的URL为: