我正在使用 Maven,Spring 3.1.1 创建一个带有模型和控制器的简单登录页面。我刚刚创建了一个模型和控制器。但是在运行应用程序时,我在浏览器和控制台中收到404 错误,我收到以下错误:
org.springframework.web.servlet.DispatcherServlet noHandlerFound
警告:在名称为“dispatcher”的 DispatcherServlet 中找不到具有 URI [/TEST2/] 的 HTTP 请求的映射
我已经正确检查了配置,但找不到确切的错误。
我改变了一些配置。我尝试/*
输入 URL-Mapping,但我面临同样的问题。
我的 Web.XML
<?xml version="1.0" encoding="ISO-8859-1" ?>
<web-app 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"
version="2.4">
<display-name>Spring MVC Application</display-name>
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/dispatcher-servlet.xml</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
</web-app>
调度程序-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" 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">
<context:component-scan base-package="com.concretepage.controller" />
<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="/pages/"/>
<property name="suffix" value=".jsp"/>
</bean>
</beans>
登录控制器.Java
package com.concretepage.controller;
import javax.servlet.http.HttpServletRequest;
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
public class LoginController {
@RequestMapping(value="/login", method = RequestMethod.GET)
public String login(){
return "redirect:pages/login.jsp";
}
@RequestMapping(value="pages/userCheck", method = RequestMethod.POST)
public String userCheck(ModelMap model, HttpServletRequest request) {
String name=request.getParameter("name");
String pwd=request.getParameter("pwd");
if("concretepage".equalsIgnoreCase(name)&&"concretepage".equalsIgnoreCase(pwd)){
model.addAttribute("message", "Successfully logged in.");
}else{
model.addAttribute("message", "Username or password is wrong.");
}
return "redirect:success.jsp";
}
}
我的错误是什么?