RequestDispatcher不使用所需的URL进行转发

时间：2015-08-01 06:18:05

标签： java spring jsp maven servlets

我从LoginPage.jsp调用LoginCheckServlet，但是当我转发页面时，我不想让URL成为LoginCheck。如果登录正确，那么我想使用/ Home转发到HomePage.jsp，如果登录不正确，我想转发到LoginRetry.jsp并使用/ LoginRetry作为url模式。

这是LoginCheckServlet Servlet。如您所见，我有@WebServlet注释（＆＃34; / LoginCheck＆＃34;）因此显示的网址为：http://localhost:8080/BankWebApp/LoginCheck 

package org.innominds.intern.BankWebApp.Servlets;

import java.io.IOException;

import javax.servlet.RequestDispatcher;
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.context.ApplicationContext;
import org.springframework.context.support.ClassPathXmlApplicationContext;
import org.springframework.jdbc.core.JdbcTemplate;
import org.springframework.context.support.AbstractApplicationContext;
import org.innominds.intern.BankWebApp.Database.*;
import org.innominds.intern.BankWebApp.Original.*;
/**
 * Servlet implementation class LoginCheckServlet
 */
@WebServlet("/LoginCheck")
public class LoginCheckServlet extends HttpServlet {
    private static final long serialVersionUID = 1L;

    /**
     * @see HttpServlet#HttpServlet()
     */
    public LoginCheckServlet() {
        super();
        // TODO Auto-generated constructor stub
    }

    /**
     * @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
     */
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        // TODO Auto-generated method stub
    }

    /**
     * @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
     */
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        //String contextPath = request.getContextPath();
        RequestDispatcher rd = null;
        ApplicationContext context = new ClassPathXmlApplicationContext("Beans.xml");
        UserJdbcTemplate jdbc = (UserJdbcTemplate)context.getBean("UserJdbcTemplate");
        String username =  request.getParameter("uname");
        System.out.println(username);
        String password = request.getParameter("password");
        User currentUser = jdbc.getUser(username);
        if(currentUser != null && currentUser.checkPassword(password)){
            request.setAttribute("currentUser", currentUser);
            //HomeServlet hs = new HomeServlet();
            //hs.doPost(request, response);
            rd = request.getRequestDispatcher("/Home"); //Calls HomeServlet which forwards to HomePage.jsp, URL should be http://localhost:8080/BankWebApp/Home
        }
        else{
            //LoginRetryServlet lr = new LoginRetryServlet();
            //lr.doPost(request, response);
            rd = request.getRequestDispatcher("/LoginRetry"); //Calls LoginRetryServlet which forwards to LoginRetry.jsp, URL should be http://localhost:8080/BankWebApp/LoginRetry
        }
        rd.forward(request,response);   
    }
}

JDBC编码没有问题，并确定用户名和密码是否正确，但我只是想更改URL模式，使其看起来很稳固。

这是我从中调用LoginCheckServlet的doPost方法的LoginPage.jsp。

<%@ 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>Login Page</title>
</head>
<body>
<h1 align = "center"> Login </h1>
<br/>
<br/>
<form action="LoginCheck" method = "post">
  <table border = "1" align = "center">
   <tr><td align = center width = 200>Username</td><td><input type="text" name= "uname"/></td></tr>
   <tr><td align = "center" width = 200>Password</td><td><input type="password" name= "password"/></td></tr>
   <!--  <tr><td></td><td align = "center"><input type="submit" value = "Login"/></td></tr> -->
  </table>
  <br/>
  <center><input align = "center" type="submit" value = "Login"/></center>
 </form>

</body>
</html>

我还没有在web.xml中完成任何映射，它基本上是空的。每次我尝试像我下面注释的那样进行映射时，我得到：localhost上的服务器Tomcat v8.0服务器无法启动。

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    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_3_0.xsd"
    id="WebApp_ID" version="3.0">
    <display-name>BankWebApp</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>WelcomePage.jsp</welcome-file>
    </welcome-file-list>
    <!-- <servlet>
        <description></description>
        <servlet-name>HomeServlet</servlet-name>
        <servlet-class>org.innominds.intern.BankWebApp.Servlets</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>HomeServlet</servlet-name>
        <url-pattern>/Home</url-pattern>
    </servlet-mapping> -->
</web-app>

有关其他信息，我使用Maven和Spring依赖项。这可能没有必要，但这是我的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>BankWebApp</groupId>
    <artifactId>BankWebApp</artifactId>
    <version>0.0.1-SNAPSHOT</version>
    <packaging>war</packaging>
    <build>
        <sourceDirectory>src</sourceDirectory>
        <plugins>
            <plugin>
                <artifactId>maven-compiler-plugin</artifactId>
                <version>3.1</version>
                <configuration>
                    <source>1.8</source>
                    <target>1.8</target>
                </configuration>
            </plugin>
            <plugin>
                <artifactId>maven-war-plugin</artifactId>
                <version>2.4</version>
                <configuration>
                    <warSourceDirectory>WebContent</warSourceDirectory>
                    <failOnMissingWebXml>false</failOnMissingWebXml>
                </configuration>
            </plugin>
        </plugins>
    </build>
    <dependencies>
        <dependency>
            <groupId>org.springframework</groupId>
            <artifactId>spring-core</artifactId>
            <version>4.1.7.RELEASE</version>
        </dependency>
        <dependency>
            <groupId>org.springframework</groupId>
            <artifactId>spring-context</artifactId>
            <version>4.1.7.RELEASE</version>
        </dependency>
        <dependency>
            <groupId>org.springframework</groupId>
            <artifactId>spring-jdbc</artifactId>
            <version>4.1.7.RELEASE</version>
        </dependency>
        <dependency>
            <groupId>mysql</groupId>
            <artifactId>mysql-connector-java</artifactId>
            <version>5.1.6</version>
        </dependency>
        <dependency>
            <groupId>commons-digester</groupId>
            <artifactId>commons-digester</artifactId>
            <version>2.1</version>
        </dependency>
        <dependency>
            <groupId>commons-logging</groupId>
            <artifactId>commons-logging</artifactId>
            <version>1.2</version>
        </dependency>

    </dependencies>
</project>

0 个答案:

没有答案
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