Question

您好我在提交以下jsp时收到了404 HTTP状态。

HTTP状态404 - / TestServlet1 你能帮我解决这个错误吗？注意：定义了人和狗类的的index.jsp

<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
    pageEncoding="ISO-8859-1"%>
<html>
<head>
</head>
<body>
<form id="a" action="/TestServlet1">
<input type="submit">
</form>
<%

%>
Name = '${person.name}'
Dog = '${person.dog.name}'
</body>
</html>

TestServlet1

package test;

import java.io.IOException;

import javax.servlet.RequestDispatcher;
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 TestServlet
 */
public class TestServlet extends HttpServlet {
    private static final long serialVersionUID = 1L;

    /**
     * @see HttpServlet#HttpServlet()
     */
    public TestServlet() {
        super();
        // TODO Auto-generated constructor stub
    }

    /**
     * @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
     */
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        // TODO Auto-generated method stub
        Dog g = new Dog();
        g.setName("dogeee");
        Person p = new Person();
        p.setDog(g);
        p.setName("xxx");
        request.setAttribute("person", p);
        RequestDispatcher dispatch = request.getRequestDispatcher("/index.jsp");
        dispatch.forward(request, response);
    }

    /**
     * @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
     */
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        // TODO Auto-generated method stub
    }

}

Web.xml中

<?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" 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">
  <display-name>GlobalWeather</display-name>
  <welcome-file-list>
    <welcome-file>index.jsp</welcome-file>
  </welcome-file-list>
  <servlet>
    <servlet-name>TestServlet</servlet-name>
    <servlet-class>test.TestServlet</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>TestServlet</servlet-name>
    <url-pattern>/TestServlet1</url-pattern>
  </servlet-mapping>
</web-app>

Answer 1

404表示找不到该网址。我怀疑您需要URL中的Web应用程序名称。

所以不要使用以下形式的动作：

/TestServlet1，

尝试

/name_of_your_web_app/TestServlet1

Answer 2

只需删除表单操作中的/：

<form id="a" action="/TestServlet1">

将其更改为

<form id="a" action="TestServlet1">

在HTML中，添加/表示相对URL而不带斜杠表示绝对URL。或者更好地使用这里提到的上下文：

<form id="a" action=${pageContext.request.contextPath}/TestServlet1>

Answer 3

你必须把你的contextroot +你的servlet名称... contextRoot通常是项目的名称。

action="/nameProject/TestServlet1"

我希望这有助于你

Answer 4

第一和第三个答案是对的。写动作=＆＃34; TestServlet1＆＃34;或action =＆＃34; / Projectname / TestServlet1＆＃34;。如果你把/ testServet1放在行动中，这意味着你正在指定所需文件的路径，在这种情况下是错误的，而如果你使用testservlet1，它意味着你正在项目中搜索文件名testservlet1来运行。

运行jsp时出现404错误

4 个答案: