我是Spring MVC的新手。在尝试基本程序时,我面临404错误页面。
Web.xml中
<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>My Demo App</display-name>
<servlet>
<servlet-name>myDemoApp</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/config/myDemoApp-servletConfig.xml</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>myDemoApp</servlet-name>
<url-pattern>*.html</url-pattern>
</servlet-mapping>
</web-app>
myDemoApp-servletConfid.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:mvc="http://www.springframework.org/schema/mvc"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:p="http://www.springframework.org/schema/p"
xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-4.0.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-4.0.xsd">
<mvc:annotation-driven/>
<!-- Components Scanner...package name should be the one which is having the controllers -->
<context:component-scan base-package="com.demo.controllers"></context:component-scan>
<!-- View Resolver Interface -->
<!-- <bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="WEB-INF/jsp/"></property>
<property name="suffix" value=".jsp"></property>
</bean> -->
<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver" p:prefix="/WEB-INF/jsp/" p:suffix=".jsp" />
</beans>
控制器类
package com.demo.controllers;
import java.util.Random;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.web.bind.annotation.RequestMapping;
@Controller
public class MyDemoController {
private String[] quotes = {"Hi I am vikram,"
+ "I am from Bangalore,"
+ "Bangalore is full of traffic"};
@RequestMapping("/getQuote")
public String getRandomQuote(Model model){
int rand = new Random().nextInt(quotes.length);
String randomQuote = quotes[rand];
//http://localhost:8080/springMVCDemo/getQuote.html
model.addAttribute("randomQuote",randomQuote);
return "quote";
}
}
JSP页面
<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%>
<!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=ISO-8859-1">
<title>My Demo App</title>
</head>
<body>
<h1>The Quote is: </h1>
<p>${randomQuote}</p>
</body>
</html>
使用的网址:
http://localhost/8050/springMVCDemo/getQuote.html
文件夹结构: [在此输入图像说明] [1]
请建议我解决错误。
答案 0 :(得分:1)
从错误中可以清楚地看出弹簧未能按照当前配置加载控制器。始终关注调试日志,并查看启动时使用弹簧容器注册的bean。记录将帮助您更好地解决此问题。
尝试使用Bootstrap监听器ContextLoaderListener
,该监听器将从contextConfigLocation
context-param
读取并加载配置
例如:
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/config/myDemoApp-servletConfid.xml</param-value>
</context-param>
<listener>
<listener-class>
org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
按照相关的post
进行操作答案 1 :(得分:0)
更改 myDemoApp (调度程序)servlet映射,如下所示。
<servlet-mapping>
<servlet-name>myDemoApp</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>