tomcat请求的资源（）不可用

小编典典

tomcat请求的资源（）不可用

tomcat

我知道这是一个非常常见的问题，因为我在包括SO这样的多个论坛中都发现了许多与此相关的问题。但是我还没有找到解决方案，但我的web.xml（位于WEB-
INF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" 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_2_5.xsd" id="WebApp_ID" version="2.5">
  <display-name>SMSProjectNew</display-name>
  <welcome-file-list>
    <welcome-file>index.html</welcome-file>
  </welcome-file-list>
  <servlet>
    <description></description>
    <display-name>ReceiveMessagesServlet</display-name>
    <servlet-name>ReceiveMessagesServlet</servlet-name>
    <servlet-class>com.sendreceive.ReceiveMessagesServlet</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>ReceiveMessagesServlet</servlet-name>
    <url-pattern>/ReceiveMessagesServlet</url-pattern>
  </servlet-mapping>
</web-app>

位于WebContent文件夹中的html页面index.html

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
The application started successfully version 1:27
<form action="/ReceiveMessagesServlet" method="post">
<input type="text" name="number"/>
<input type="text" name="message"/>
<input type="submit" name="submit"/>
</form> 
</body>
</html>

最后，位于src \ com.sendreceive包com.sendreceive包中的Servlet ReceiveMessagesServlet；

import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

   public class ReceiveMessagesServlet extends HttpServlet {
    private static final long serialVersionUID = 1L;

    public ReceiveMessagesServlet() {
        super();
        // TODO Auto-generated constructor stub
    }


    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        processRequest(request,response);
    }

    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        processRequest(request,response);
    }

    protected void processRequest(HttpServletRequest request,
            HttpServletResponse response) {
        String responseMessage = request.getParameter("message");
        String responseNumber = request.getParameter("number");
        System.out.println(responseMessage+responseNumber);
    }

}

我已经在Eclipse中安装了tomcat插件。当我右键单击该项目，然后单击在服务器上运行该项目。tomcat服务器从eclipse启动，并显示index.html页面。.但是当我在字段中输入一些值并单击Submit..it时出现404错误。.i过去2个小时一直在苦苦挣扎。也..fyi，我正在使用本教程
 http://www.ibm.com/developerworks/opensource/library/os-eclipse-
tomcat/index.html

阅读 2283

2020-06-16

共1个答案

小编典典

由于action =“ / ReceiveMessagesServlet”，您收到404错误，请删除斜杠。尝试使用action =“
ReceiveMessagesServlet”。

当您在URL模式中添加斜杠时，容器将查找名称为’ReceiveMessagesServlet’的Web应用程序部署，因为此位置不存在，您将收到404错误。

2020-06-16