0

在这里学习一些 Java Web 开发并尝试启动我的第一个 servlet。但我收到此错误： HTTP method GET is not supported by this URL

description The specified HTTP method is not allowed for the requested resource (HTTP method GET is not supported by this URL).

我在我的 html 表单中使用了“get”方法，如您所见，我已经实现了 doget 方法。但我不确定为什么会收到此错误。会不会和我的有关系web.xml？

我尝试通过将 html 方法更改为“post”并使用 doPost 方法来使用 POST 方法，但我也得到了等效的错误。我这样做只是为了测试，我不想使用 post。

网页.xml：

<web-app 
    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/web-app_2_4.xsd"
    version="2.4">
  <servlet>
     <servlet-name>AppleFox</servlet-name>
     <servlet-class>com.AppleFox.web.ProcessQuery</servlet-class>
  </servlet>

  <servlet-mapping>
     <servlet-name>AppleFox</servlet-name>
     <url-pattern>/ProcessQuery.do</url-pattern>
  </servlet-mapping>
</web-app>

小服务程序代码：

package com.AppleFox.web;

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

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

/**
 * Servlet implementation class ProcessQuery
 */
public class ProcessQuery<HttpServletRequest> 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 {
        response.setContentType("text/html");
        PrintWriter out = response.getWriter();
        String query = ((ServletRequest) request).getParameter("query");

        out.println("Sorry we could find any results for " + query + ".");
    }

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

我还在我的 html 表单中使用了“get”方法。不知道为什么我会收到此错误。

4

2 回答 2

2

只需更改以下行并重新排列导入，一切都会正常工作：

改变：

public class ProcessQuery<HttpServletRequest> extends HttpServlet {

到

public class ProcessQuery extends HttpServlet {

将导入更改为：

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

import javax.servlet.ServletException;
import javax.servlet.ServletRequest;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
于 2012-06-26T22:48:04.583 回答
0

如果您有兴趣同时使用 doget 和 dopost 方法，您可以将操作 java 代码放入 proccessRequest(request,response) 并在 dopost 和 doget 方法中调用它，如下所示

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
package controller;

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;

/**
 *
 * @author Birhan
 */
@WebServlet(name = "servlet", urlPatterns = {"/servlet"})
public class servlet extends HttpServlet {

    /**
     * Processes requests for both HTTP <code>GET</code> and <code>POST</code>
     * methods.
     *
     * @param request servlet request
     * @param response servlet response
     * @throws ServletException if a servlet-specific error occurs
     * @throws IOException if an I/O error occurs
     */
    protected void processRequest(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        response.setContentType("text/html;charset=UTF-8");
        try (PrintWriter out = response.getWriter()) {
            /* TODO output your page here. You may use following sample code. */
            out.println("<!DOCTYPE html>");
            out.println("<html>");
            out.println("<head>");
            out.println("<title>Servlet servlet</title>");            
            out.println("</head>");
            out.println("<body>");
            out.println("<h1>Servlet servlet at " + request.getContextPath() + "</h1>");
            out.println("</body>");
            out.println("</html>");
        }
    }

    // <editor-fold defaultstate="collapsed" desc="HttpServlet methods. Click on the + sign on the left to edit the code.">
    /**
     * Handles the HTTP <code>GET</code> method.
     *
     * @param request servlet request
     * @param response servlet response
     * @throws ServletException if a servlet-specific error occurs
     * @throws IOException if an I/O error occurs
     */
    @Override
    protected void doGet(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        processRequest(request, response);
    }

    /**
     * Handles the HTTP <code>POST</code> method.
     *
     * @param request servlet request
     * @param response servlet response
     * @throws ServletException if a servlet-specific error occurs
     * @throws IOException if an I/O error occurs
     */
    @Override
    protected void doPost(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        processRequest(request, response);
    }

    /**
     * Returns a short description of the servlet.
     *
     * @return a String containing servlet description
     */
    @Override
    public String getServletInfo() {
        return "Short description";
    }// </editor-fold>

}
于 2017-01-03T08:40:55.623 回答