Question

我创建了dynamic web project，并添加了两项：

index.jsp这样的网页：

<%@ 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="GrettingServlet" method="POST">
        First Name: <input type="text" name="firstName" size="20"><br>
        Last Name: <input type="text" name="lastName" size="20">
        <br><br>
        <input type="submit" value="Submit">
</form> 

</body>
</html>

默认包 servlet就像这样（名为GrettingServlet.java）：

import java.io.IOException;
import java.io.PrintWriter;

import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

public class GrettingServlet extends HttpServlet {
    private static final long serialVersionUID = 1L;

    public GrettingServlet() {
        super();
    }

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

    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        response.setContentType("text/html;charset=UTF-8");
        PrintWriter out = response.getWriter();
        String firstName = request.getParameter("firstName").toString();
        String lastName = request.getParameter("lastName").toString();

        out.println("<html>");
        out.println("<head>");
        out.println("<title>Servlet GreetingServlet</title>");
        out.println("</head>");
        out.println("<body>");
        out.println("<p>Welcome " + firstName + " " + lastName + "</p>");
        out.println("</body>");
        out.println("</html>");

        out.close();
    }

}

我安装了tomcat6，因此我有Apache Software Foundation个文件夹。最后，我想创建此项目的war文件，因此我选择了项目Export>War file，并在Destination文本中选择了路径{{1}中的webapps文件夹}。该项目名为C:\Program Files (x86)\Apache Software Foundation\Tomcat 6.0\webapps。并且为了在服务器上看到MyFirstServlet的形式，我在浏览器index.jsp中写了但是我得到了消息

http://localhost:8080/MyFirstServlet/

servlet映射是这样的：

HTTP Status 404 - /MyFirstServlet/

type Status report

message /MyFirstServlet/

description The requested resource (/MyFirstServlet/) is not available.

Apache Tomcat/6.0.35

我在<?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" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5"> <display-name>MyFirstServlet</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>default.jsp</welcome-file> </welcome-file-list> <servlet> <description>new</description> <display-name>GrettingServlet</display-name> <servlet-name>GrettingServlet</servlet-name> <servlet-class>GrettingServlet</servlet-class> </servlet> <servlet-mapping> <servlet-name>GrettingServlet</servlet-name> <url-pattern>/GrettingServlet</url-pattern> </servlet-mapping> </web-app>

可能是什么问题？

Answer 1

从给定的示例中，您需要将Web应用程序部署到Tomcat中作为MyFirstServlet.war（或作为展开的目录 - 这没有区别）并将GrettingServlet映射到应用程序根 - 如果您希望servlet处理根目录：

您的/WEB-INF/web.xml应该这样：

<servlet>
    <servlet-name>GrettingServlet</servlet-name>
    <servlet-class>your.package.GrettingServlet</servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>GrettingServlet</servlet-name>
    <url-pattern>/</url-pattern>
</servlet-mapping>

请注意拼写错误：“Gretting”（在servlet和映射中）与“Greeting”（以JSP形式）

使用您的设置，您应该将浏览器指向http://localhost:8080/MyFirstServlet/GrettingServlet以访问servlet。

如果您的想法是让JSP页面来处理根目录，那么您应该浏览http://localhost:8080/MyFirstServlet/<yourJSPName>.jsp或者使用名为index.jsp或default.jsp的JSP（请参阅<welcome-file-list/>你web.xml的一部分。在这种情况下，我想，您的想法是显示JSP然后发布到servlet，因此请确保您的servlet规范和映射是正确的（web.xml servlet映射和JSP表单action属性）

Answer 2

只需使用以下内容修改您的web.xml即可

<servlet>
  <servlet-name>GrettingServlet</servlet-name>
  <servlet-class>GrettingServlet</servlet-class>
</servlet>
<servlet-mapping>
  <servlet-name>GrettingServlet</servlet-name>
  <url-pattern>/GreetingServlet</url-pattern>
</servlet-mapping>

更新这里是整个WEB.XML

<?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" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
  <display-name>MyFirstServlet</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>default.jsp</welcome-file>
  </welcome-file-list>
  <servlet>
    <description>new</description>
    <display-name>GrettingServlet</display-name>
    <servlet-name>GrettingServlet</servlet-name>
    <servlet-class>GrettingServlet</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>GrettingServlet</servlet-name>
    <url-pattern>/GreetingServlet</url-pattern>
  </servlet-mapping>
</web-app>

用tomcat 6运行jsp文件

2 个答案: