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Servlets 当我运行以下代码时,我收到一个Servlet 404错误_Servlets_Http Status Code 404 - Fatal编程技术网

Servlets 当我运行以下代码时,我收到一个Servlet 404错误

Servlets 当我运行以下代码时,我收到一个Servlet 404错误,servlets,http-status-code-404,Servlets,Http Status Code 404,我正在运行以下代码,这是html版本,它显示了姓名和年龄框 第一页 输入您的姓名: 输入您的年龄: 但是,当我运行相同逻辑的代码(servlet版本)时,它得到了一个错误 The origin server did not find a current representation for the target resource or is not willing to disclose that one exists. 代码如下: import java.io.IOException;

我正在运行以下代码,这是html版本,它显示了姓名和年龄框


第一页
输入您的姓名:
输入您的年龄:
但是,当我运行相同逻辑的代码(servlet版本)时,它得到了一个错误

The origin server did not find a current representation for the target resource or is not willing to disclose that one exists. 
代码如下:

import java.io.IOException;
导入javax.servlet.ServletException;
导入javax.servlet.annotation.WebServlet;
导入javax.servlet.http.HttpServlet;
导入javax.servlet.http.HttpServletRequest;
导入javax.servlet.http.HttpServletResponse;
导入java.io.*;
@WebServlet(“/Main”)
公共类Main扩展了HttpServlet{
私有静态最终长serialVersionUID=1L;
公用干管(){
超级();
}
受保护的无效数据集(HttpServletRequest请求,HttpServletResponse
响应)抛出ServletException、IOException{
PrintWriter out=response.getWriter();//已添加
out.println(“”);

println(“输入您的名字:确保在servlet中添加所有html元素(标记)

试试这个:

  protected void doGet(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        PrintWriter out = response.getWriter();     //Is added

        String htmlTag
                = "<!doctype html public \"-//w3c//dtd html 4.0 " + "transitional//en\">\n";

        out.println(htmlTag
                + "<html>\n"
                + "<head><title>Reg Form</title></head>\n"
                + "<body bgcolor = \"#f0f0f0\">\n"
                + "<h1 align = \"center\">Regform</h1>\n"
                + "<ul>\n"
                + "<form action=\"http://localhost\" method=\"post\">"
                + "Enter Your Name: <input type=\"text\" name = \"yourName\"</input><br><br>\n"
                + "Enter Your Age&nbsp;&nbsp; : <input type=\"text\" name = \"yourAge\" </input>\n"
                + "</ul>\n"
                + "</body>"
                + "</html>");

        out.close();
}
protectedvoid doGet(HttpServletRequest请求,HttpServletResponse响应)
抛出ServletException、IOException{
PrintWriter out=response.getWriter();//已添加
字符串htmlTag
=“\n”;
out.println(htmlTag
+“\n”
+“注册表单\n”
+“\n”
+“Regform\n”
+“
    \n” + ""
    + "输入您的姓名:您点击的URL是什么以获得响应?尝试点击我建议您了解servlet的工作原理。您缺少的一点是servlet的基本功能。您使用URL的方式不正确。此外,您尝试从类WEB-INF/类内部访问Main.java,该类表示您需要使用它r Java基础。基本上,一旦编译Java类,就会创建一个包含字节码的.class文件。在WAR文件中,它将是WEB-INF/classes/main.class,但您不能直接访问它。您需要访问为servlet指定的URL。您能给我看一下您的WEB.xml吗?abc yi.main
      protected void doGet(HttpServletRequest request, HttpServletResponse response)
                throws ServletException, IOException {
            PrintWriter out = response.getWriter();     //Is added
    
            String htmlTag
                    = "<!doctype html public \"-//w3c//dtd html 4.0 " + "transitional//en\">\n";
    
            out.println(htmlTag
                    + "<html>\n"
                    + "<head><title>Reg Form</title></head>\n"
                    + "<body bgcolor = \"#f0f0f0\">\n"
                    + "<h1 align = \"center\">Regform</h1>\n"
                    + "<ul>\n"
                    + "<form action=\"http://localhost\" method=\"post\">"
                    + "Enter Your Name: <input type=\"text\" name = \"yourName\"</input><br><br>\n"
                    + "Enter Your Age&nbsp;&nbsp; : <input type=\"text\" name = \"yourAge\" </input>\n"
                    + "</ul>\n"
                    + "</body>"
                    + "</html>");
    
            out.close();
    }
    
      <servlet>
            <servlet-name>Main</servlet-name>
            <servlet-class>test.Main</servlet-class>  //test is the Package
        </servlet>
        <servlet-mapping>
            <servlet-name>Main</servlet-name>
            <url-pattern>/Main</url-pattern>
        </servlet-mapping>