基于xml配置Servlet

1.Java代码部分

public class TestXmlServlet extends HttpServlet {
    @Override
    protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
        super.doPost(req, resp);
    }

    @Override
    protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
        resp.setContentType("text/html;charset=utf-8");
        resp.getWriter().append("访问成功");
    }
}

2.web.xml配置
 

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_4_0.xsd"
         version="4.0">
    <!-- servlet的配置 -->
    <servlet>
        <!-- 给servlet起个名字，没有太大的实际作用 -->
        <display-name>TestXmlServlet</display-name>
        <!-- servlet-name 要和servlet-mapping里面的servlet-name对应 -->
        <servlet-name>TestXmlServlet</servlet-name>
        <!-- 这里配置对应的类， -->
        <servlet-class>com.dl.code.TestXmlServlet</servlet-class>
        <!-- tomcat一启动就加载 -->
        <load-on-startup>1</load-on-startup>
    </servlet>
    <!-- servlet的映射 -->
    <servlet-mapping>
        <servlet-name>TestXmlServlet</servlet-name>
        <!-- 访问的路径 -->
        <url-pattern>/testXmlServlet.do</url-pattern>
    </servlet-mapping>
</web-app>