Javaservlet RandomAccessFile 错误答案

【问题标题】：Javaservlet RandomAccessFile errorsJavaservlet RandomAccessFile 错误
【发布时间】：2016-09-24 23:04:57
【问题描述】：

我收到错误“HTTP 状态 404-/servlet/FormPost 类型状态报告消息 /servlet/FormPost 描述请求的资源不可用 Apache Tomcat/7.0.72"

谁能看到我的代码有什么问题？我正在使用 Eclipse 和 Tomcat v7.0。我试图让程序输出一个表单，写入一个随机访问文件，并从随机访问文件返回结果。我是一名 SQL 程序员，不是 Java servlet 的粉丝，也不是很了解它们。请解释一下，就好像我是编程新手一样，或者编辑我的代码。谢谢！

这是我的代码：

package net.codejava;

import java.io.*;
import javax.servlet.*;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.*;

@WebServlet(name="FormPost", urlPatterns={"/FormPost"})

public class FormPost extends HttpServlet   {

   private static final long serialVersionUID = 1L;

   private String target = "localhost:7070";

   public void doGet(HttpServletRequest request,HttpServletResponse response) throws IOException, ServletException{
       response.setContentType("text/html");
       PrintWriter out = response.getWriter();

       printHeader(out);
       printForm(out);
       printFooter(out);
   }

     public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException{
         final int BUFFER_LENGTH = 41;
         String data = null;
         PrintWriter out = response.getWriter();
         RandomAccessFile randomAccessFile = new RandomAccessFile("C:/temp/week4.dat", "rw");

         StringBuffer buffer = new StringBuffer(request.getParameter("myName"));
         StringBuffer buffer2 = new StringBuffer(request.getParameter("myAddress"));
         StringBuffer buffer3 = new StringBuffer(request.getParameter("myCityAndState"));
         randomAccessFile.seek(randomAccessFile.length());
         randomAccessFile.writeBytes(buffer.toString());
         randomAccessFile.seek(0);
         printForm(out);

         buffer.setLength(BUFFER_LENGTH);
         buffer.setCharAt(BUFFER_LENGTH - 1, '\n');

         randomAccessFile.seek(randomAccessFile.length());
         randomAccessFile.writeChars(buffer.toString());
         randomAccessFile.writeChars(buffer2.toString());
         randomAccessFile.writeChars(buffer3.toString());
         randomAccessFile.seek(0);

         while ((data = randomAccessFile.readLine()) != null){
             out.println(data + "<br />");
             randomAccessFile.close();
             doPost(request, response);
         }
         printFooter(out);
     }

     public void printHeader(PrintWriter out){
         out.println("<!DOCTYPE html>");
         out.println("<html lang='en'>");
         out.println("<title>");
         out.println("Servlet Read Write File");
         out.println("</title>");
         out.println("<body>");
         out.println("<div>");
     }

     public void printFooter(PrintWriter out){
         out.println("</div>");
         out.println("</body>");
         out.println("</html>");
     }

     public void printForm(PrintWriter out){
         out.println("<form method='post' action='http://" + target + "/MyServlet/FormPost'>");
         out.println("<label>Enter Name</label>&nbsp;&nbsp;&nbsp;");
         out.println("<input type='text' name='myName' size='30' maxlength='30'/>");
         out.println("<input type='submit' />");
         out.println("<label>Enter Address</label>&nbsp;&nbsp;&nbsp;");
         out.println("<input type='text' name='myAddress' size='30' maxlength='30'/>");
         out.println("<input type='submit' />");
         out.println("<label>Enter City and State</label>&nbsp;&nbsp;&nbsp;");
         out.println("<input type='text' name='myCityAndState' size='30' maxlength='30'/>");
         out.println("<input type='submit' />");
         out.println("</form>");
     }
}

【问题讨论】：

标签： java eclipse tomcat servlets

【解决方案1】：

该问题与您的方法实现无关。
HTTP 404 告诉您 tomcat http-server 找不到您的请求的任何处理程序。在一个简单的情况下，您的请求应该类似于 "localhost:8080/<your_war_file_/_app_name>/<string_matching_your_web_servlet_pattern>"

【讨论】：

谢谢，这是一个简单的错误，在我的 printForm 中，我更正了这个问题，但现在在下面收到“at net.codejava.FormPost.doPost(FormPost.java:56)”线。 doPost（请求，响应）；
@Natela 你能接受吗，因为我是新手，我看到这样一个不被接受的答案会降低声誉。