Servlet如何将控制权转交给另一个页面

Question

[code]package com.zxm.web;import java.io.IOException;
import java.io.PrintWriter;import javax.servlet.RequestDispatcher;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;public class CompIntrDeal extends HttpServlet { public CompIntrDeal() {
super();
}
public void destroy() {
super.destroy(); 
}

public void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
RequestDispatcher rd = request.getRequestDispatcher("/admin/header.html");
rd.forward(request, response);
} public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
doGet(request,response);
} public void init() throws ServletException {
}}
[/code]本来预期结果是跳转到header.html页面。
但浏览器中却显示了下面的一条消息，实际上转到的并不是head.html页面：

Answer 1

我的servlet源代码是这样的：package com.zxm.web;import java.io.IOException;
import java.io.PrintWriter;import javax.servlet.RequestDispatcher;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;public class CompIntrDeal extends HttpServlet { public CompIntrDeal() {
super();
}
public void destroy() {
super.destroy(); 
}

public void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
RequestDispatcher rd = request.getRequestDispatcher("/admin/header.html");
rd.forward(request, response);
} public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
doGet(request,response);
} public void init() throws ServletException {
}}

Answer 2

java技术交流群100756746，希望大家加入，我们共同讨论，共同进步！

Answer 3

form的action是怎么写的，
web.xml里servlet映射好了吗？

Answer 4

/admin/compIntr.jsp中form表单的部分内容：
out.println("<form action=\"CompIntrDeal\" method=\"post\">");
web.xml的servlet映射：  <servlet>
    <description>This is the description of my J2EE component</description>
    <display-name>This is the display name of my J2EE component</display-name>
    <servlet-name>CompIntrDeal</servlet-name>
    <servlet-class>com.zxm.web.CompIntrDeal</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>CompIntrDeal</servlet-name>
    <url-pattern>/admin/CompIntrDeal</url-pattern>
  </servlet-mapping>映射应该是没有什么问题的，都是通过myeclipse自动产生的

Answer 5

如果Servlet配置没有问题的话，
可能是
RequestDispatcher rd = request.getRequestDispatcher("/admin/header.html");
这个地方跳转页面路径的问题
改成：RequestDispatcher rd = request.getRequestDispatcher("header.html");
试试看....

Answer 6

<?xml version="1.0" encoding="UTF-8"?>
<web-app 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">
 <context-param>
  <param-name>host</param-name>
  <param-value>localhost</param-value>
 </context-param>
 <context-param>
  <param-name>database</param-name>
  <param-value>website</param-value>
 </context-param>
 <context-param>
  <param-name>user</param-name>
  <param-value>root</param-value>
 </context-param>
 <context-param>
  <param-name>password</param-name>
  <param-value>123</param-value>
 </context-param>
 <context-param>
  <param-name>JDBCDriver</param-name>
  <param-value>com.mysql.jdbc.Driver</param-value>
 </context-param>
 <context-param>
  <param-name>dbPort</param-name>
  <param-value>3306</param-value>
 </context-param>
  <servlet>
    <description>This is the description of my J2EE component</description>
    <display-name>This is the display name of my J2EE component</display-name>
    <servlet-name>CompIntrDeal</servlet-name>
    <servlet-class>com.zxm.web.CompIntrDeal</servlet-class>
  </servlet>  <servlet-mapping>
    <servlet-name>CompIntrDeal</servlet-name>
    <url-pattern>/admin/CompIntrDeal</url-pattern>
  </servlet-mapping>
 <welcome-file-list>
  <welcome-file>index.jsp</welcome-file>
 </welcome-file-list>
 <login-config>
  <auth-method>BASIC</auth-method>
 </login-config>
</web-app>

Answer 7

    public void doGet(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        RequestDispatcher rd = request.getRequestDispatcher("/admin/header.html");
        rd.forward(request, response);
    }    public void doPost(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        doGet(request,response);
    }换一下试试看：    public void doGet(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
               doPost(request,response);    }    public void doPost(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
          RequestDispatcher rd = request.getRequestDispatcher("/admin/header.html");
          rd.forward(request, response);    }

Answer 8

RequestDispatcher rd = request.getRequestDispatcher("/admin/header.html"); 
rd.forward(request, response); 

Answer 9

估计是 那个servlet的mapping没写好