Question

我正在编写一些基本的servlet，并且我在链接http://localhost:8080/Parameters/wypisz_parametry?parametr1=java&parametr2=start&parametr3=javaEE中传递参数，但是当我输入它时，我得到404错误。

的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_3_0.xsd" id="WebApp_ID" version="3.0">
  <servlet>
    <servlet-name>Parameters</servlet-name>
    <servlet-class>pl.javastart.servlets.wypisz_parametry.java</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>Parameters</servlet-name>
    <url-pattern>/wypisz_parametry</url-pattern>
  </servlet-mapping>
</web-app>

wypisz_parametry类：

package pl.javastart.servlets;

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 wypisz_parametry extends HttpServlet {
private static final long serialVersionUID = 1321321321L;

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

  String parametr1 = request.getParameter("parametr1");
  String parametr2 = request.getParameter("parametr2");
  String parametr3 = request.getParameter("parametr3");

  out.println("Wczytanie 3 parametrow z zadania :");
  out.println(parametr1);
  out.println(parametr2);
  out.println(parametr3);
  }
 }

project structure

Answer 1

错误在web.xml中，类名应该没有.java

<servlet-class>pl.javastart.servlets.wypisz_parametry</servlet-class>

链接中的passin参数时出现错误404

1 个答案: