Question

我制作了一个简单的tomcat servlet，完全遵循Youtube tutorial。

我创建了一个名为XmlServlet的类，它处理简单的GET和POST请求。

这是我的班级：

首先包装：

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 XmlServlet extends HttpServlet {
    protected void doGet(HttpServletRequest request,
            HttpServletResponse response) throws ServletException, IOException {
        response.setContentType("text/html");
        PrintWriter out = response.getWriter();
        String UserName = request.getParameter("UserName");
        String FullName = request.getParameter("FullName");

        System.out.println("hello u" + UserName);
    }

    protected void doPost(HttpServletRequest request,
            HttpServletResponse response) throws ServletException, IOException {
        response.setContentType("text/html");
        PrintWriter out = response.getWriter();
        String UserName = request.getParameter("UserName");
        String FullName = request.getParameter("FullName");
        String location = request.getParameter("location");
        String prof = request.getParameter("prof");
        out.println("hello u from dpost" + UserName + "u full name" + FullName
                + "address" + location);

    }

}

和Servlet：

package First;

import java.io.IOException;
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 SimpleServlet
 */
@WebServlet(description = "My First One", urlPatterns = { "/SimpleServletPath" })
public class SimpleServlet extends HttpServlet {
    private static final long serialVersionUID = 1L;

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

    /**
     * @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
     */
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {


    }

}

和我的web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns="http://xmlns.jcp.org/xml/ns/javaee"
    xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
    id="WebApp_ID" version="3.1">
    <!--  
    <servlet>
        <servlet-name>XmlServlet</servlet-name>
        <servlet-class>First.Xmlservlet</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>XmlServlet</servlet-name>
<url-pattern>/SimpleServletPath</url-pattern>
    </servlet-mapping>
-->
</web-app>



and i create simple html page to represent this 
<!DOCTYPE html>
<html>
<head>
<meta charset="ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
    <form method="get" action="SimpleServletPath">

        user name<input name="UserName" /> full name<input name="FullName" />
        <br>
         Work <input type="radio" name="prof" value="developer">developer</input>
        <input type="radio" name="prof" value="IT">IT</input>
         <select name="location" multiple size=3 >
            <option value="here">here</option>
            <option value="home">home</option>
            <option value="school">school</option>
            <option value="football">football</option>
            <option value="chelsea">Chelsea</option>
        </select> 
        </br> 
        <input type="Submit" />
    </form>
</body>
</html>

我无法从HTML页面获取任何数据。一切似乎都很好，但我收到有关web.xml的错误。有人可以帮助检查我的servlet，类和web.xml并告诉我问题所在吗？

Answer 1

因此，当您在HTML页面上提交并提交时，它将转到servlet并获得一个空白页面，对吧？好吧，那是因为你的servlet在doGet方法中几乎没有任何东西。请注意，您的表单action为SimpleServletPath而不是XmlServlet。但等等，它变得更糟！在SimpleServlet java文件中，您将带有注释的路径映射到SimpleServletPath。但是，您还将第二个servlet XmlServlet映射到与web.xml相同的路径（！）....所以它一团糟。（由于双重映射，您实际上可能会得到一个错误页面。）

因此，要解决此问题，请将每个servlet映射到其自己的URL，而不是同一URL。然后确保您从HTML表单中实际提交的任何servlet实际上都会在响应中打印一些内容。决定是否要将servlet映射到带有注释的URLS或使用web.xml而不是混合使用这两种方法也是一个好主意。

使用tomcat server 8.0的web.xml问题

1 个答案: