我是JSP和Servlets的新手,
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<servlet>
<servlet-name>LoginPage</servlet-name>
<servlet-class>com.planner.servlet.LoginPage</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>LoginPage</servlet-name>
<url-pattern>/LoginPage</url-pattern>
</servlet-mapping>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/applicationContextBL.xml</param-value>
</context-param>
</web-app>
<%@ 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 method="POST" action="/LoginPage">
<table border="1">
<tr>
<td>Login</td>
<td><input type="text" name="login" />
</td>
</tr>
<tr>
<td>Senha:</td>
<td><input type="password" name="pass" /></td>
</tr>
<tr>
<input type="submit" value="Entrar" />
</tr>
</table>
</form>
</body>
</html>
package com.planner.servlet;
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.springframework.beans.factory.config.AutowireCapableBeanFactory;
import org.springframework.web.context.WebApplicationContext;
import org.springframework.web.context.support.WebApplicationContextUtils;
/**
* Servlet implementation class LoginPage
*/
public class LoginPage extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public LoginPage() {
super();
}
@Override
public void init() throws ServletException {
WebApplicationContext ctx = WebApplicationContextUtils.getRequiredWebApplicationContext(getServletContext());
AutowireCapableBeanFactory bf = ctx.getAutowireCapableBeanFactory();
// bf.autowireBean(this);
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse
* response)
*/
protected void service(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException {
PrintWriter out = response.getWriter();
out.println("<html>");
out.println("<head>");
out.println("<title> TESTE </title>");
out.println("</head>");
out.println("<body>");
out.println("Hello World!");
out.println("</body>");
out.println("</html>");
}
}
问题: 当我点击提交按钮时,它会显示错误:
**type Status report
message /Apontador_Web/LoginPage
description The requested resource is not available.**
当我删除标签“listener”和“context-param”时 它加载servlet。
可能会发生什么?
答案 0 :(得分:1)
转到 glassfish服务器输出并查找一些异常
告诉我例外
的名称全部打印
答案 1 :(得分:0)
您需要在类名@WebServlet(urlPatterns={"/LoginPage"})
public class LoginPage extends HttpServlet {