我使用弹簧控制器注释。 但运行我的应用程序获取HTTP状态404时 -/LoginAuth错误HTTP状态404 -/LoginAuth
我的文件点击登录按钮,它会调用的LoginController后 的web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
id="WebApp_ID"
version="2.5"
xmlns="http://java.sun.com/xml/ns/javaee"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" >
<display-name>SpringTest</display-name>
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
<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>*.do</url-pattern>
</servlet-mapping>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>WEB-INF/dispatcher-servlet.xml</param-value>
</context-param>
</web-app>
的index.jsp 在这个文件中。
<%@ 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>Insert title here</title>
</head>
<body>
<form action="LoginAuth" method="get">
<label>Enter name :</label>
<input type="text" name="uname"><br>
<label>Enter pass :</label>
<input type="password" name="pass"><br>
<input type="submit" value="LogIn">
</form>
</body>
</html>
dispacher-servlet.xml中
<beans xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
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">
<bean
id="viewResolver"
class="org.springframework.web.servlet.view.InternalResourceViewResolver" >
<property
name="prefix"
value="/WEB-INF/JSP/" />
<property
name="suffix"
value=".jsp" />
</bean>
<context:component-scan base-package="com.controller" />
</beans>
LoginController.java
package com.controller;
import org.apache.catalina.connector.Request;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.servlet.ModelAndView;
@Controller
public class LoginController {
@RequestMapping(value="/LoginAuth",method=RequestMethod.GET)
public ModelAndView loginAuth(Request req){
String uname=req.getParameter("uname");
String pass=req.getParameter("pass");
ModelAndView mav=new ModelAndView();
if(uname.equals("Alk") && pass.equals("Alk1234")){
mav.setViewName("success");
}else{
mav.setViewName("error");
}
return mav;
}
}
请帮助我。
是不是因为您将调度程序servlet映射到***。do **和**/LoginAuth **只是与此url模式不匹配? – oceansize