我该如何解决这个问题--->下面的代码中出现错误 HTTP 状态 404答案

【问题标题】：How can i address this---> Error HTTP Status 404 on my Code below我该如何解决这个问题--->下面的代码中出现错误 HTTP 状态 404
【发布时间】：2018-01-28 14:49:39
【问题描述】：

所以我正在尝试使用 servlet 在 Web 中显示信息，但我被卡住了，我尝试搜索和研究但徒劳无功，我没有找到任何帮助。我想知道这可能是服务器的问题吗？当我尝试使用 Java 中的 dopost 方法运行我的 html 时，它一直给我这个错误HTTP Status 404。否则，当我在服务器上运行我的 Java 类时，它显示没有任何问题，当我在网络上运行我的 html 时，它显示没有任何问题，但 CANNOT 获取方法。请任何帮助将不胜感激。我的代码如下.. /我的.JSP 文件也出现同样的错误。这只是我决定使用的示例，以解决我的主要代码问题。

我的 Java 代码

package helloServelets;

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;

/**
 * Servlet implementation class LearningServelets
 */
@WebServlet("/LearningServelets")
public class LearningServelets extends HttpServlet {


    /**
     * 
     */
    private static final long serialVersionUID = 1L;

    /**
     * @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
     */
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    String yourname = request.getParameter("YourName");
    String enter = request.getParameter("Enter");
    String school = request.getParameter("SchoolName");
    String work = request.getParameter("Work");
    String Home = request.getParameter("home");


        response.setContentType("text/html");

        PrintWriter output = response.getWriter();

        output.println("<html><body><h3>Hello " + yourname);

        output.println("</h3><br />" + enter + " + " + school+work);
        output.println(Home + "<br /> "  + "</body></html>");

    }

    /**
     * @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
     */
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        doGet(request,response);
    }

}

我的HTML代码如下

<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Insert title here</title>
</head>
<body>
<form method="post" action="LearningServelets">

What is your YourName?
<input name = "Yourname"> 
<p>
What is your Gender

<input name ="Enter">
</p>
<p>
Where did you go to school?
<input name = "SchoolName">
</p>

<p>
Where do you Work?
<input name = "Work">
</p>

<p>
Where do you live?
<input name = "home">
</p>
<input type ="submit">
</form>

</body>
</html>

而我的XML文件如下

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd" id="WebApp_ID" version="3.1">
  <display-name>ServeletsLearning</display-name>
      <servlet> 
     <servlet-name>LearningServelets</servlet-name>
     <servlet-class>helloServelets.LearningServelets</servlet-class>
     </servlet>

</web-app>

【问题讨论】：

404 是来自 helloServelets 的 http 错误。您的帖子标题具有误导性，因为该问题与 XML 无关。

标签： java html xml tomcat

【解决方案1】：

试试这个：

index.html 文件

<!DOCTYPE html>
<html>
    <head>
        <meta charset="UTF-8">
        <title>Insert title here</title>
    </head>
    <body>
    <form method="post" action="LearningServlet">

        What is your age?<input name = "age"><br> 
        What is your Gender?<input name ="gender"><br>
        Where did you go to school?<input name = "schoolname"><br>          
        Where do you Work?<input name = "work"><br>
        Where do you live?<input name = "home"><br>
        <input type ="submit">
    </form> 
</body>

LearningServlet.java 文件打包 helloservlet;

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;
@WebServlet("/LearningServlet")
public class LearningServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
public LearningServlet() {
    super();
}
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
}
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    String age=request.getParameter("age");
    String gender=request.getParameter("gender");
    String schoolname=request.getParameter("schoolName");
    String work=request.getParameter("work");
    String home=request.getParameter("home");

    response.setContentType(getServletContext().getMimeType("text/html"));
    PrintWriter pw=response.getWriter();
    pw.write("<html><body>");
    pw.write("Age           :"+age+"<br>");
    pw.write("Gender        :"+gender+"<br>");
    pw.write("School Name   :"+schoolname+"<br>");
    pw.write("Work          :"+work+"<br>");
    pw.write("Home          :"+home+"<br>");
    pw.write("</body></html>");
   }
}

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd" version="3.1">
  <display-name>helloservlet</display-name>
  <welcome-file-list>
    <welcome-file>index.html</welcome-file>
  </welcome-file-list>
</web-app>

现在在服务器上部署您的项目并在浏览器中打开： http://localhost:8080/helloservlet/index.html

注意：假设您的应用程序/项目名称是 helloservlet

【讨论】：

你的意思是我删除了 XML 中的所有内容？我会继续使用 ServeletsLearning 吗？您可以尝试发布已删除的代码，我将不胜感激
嗯，我被困了，想继续前进。这让我很紧张。
我已经编辑了我的答案。进行必要的更改或只是复制粘贴代码。让我们知道它是否适合您。快乐编码。
谢谢让我看看。 :)