javaWeb之写一个最简单的servlet

1. 创建一个类servletTest2 继承HttpServlet类。

public class servletTest2 extends HttpServlet {

    public servletTest2(){
        super();
    }

    @Override
    protected void doGet(HttpServletRequest req, HttpServletResponse resp)
            throws ServletException, IOException {

        resp.setContentType("text/html");
        PrintWriter out = resp.getWriter();
        out.println("<!DOCTYPE HTML PUBLIC \"-//W3C//DTD HTML 4.01 Transitional//EN\">");
        out.println("<HTML>");
        out.println("  <HEAD><TITLE>A Servlet</TITLE></HEAD>");
        out.println("  <BODY>");
        out.print("    This is ");
        out.print(this.getClass());
        out.println(", using the GET method");
        out.println("  </BODY>");
        out.println("</HTML>");
        out.flush();
        out.close();
    }
    
}

web.xml中增加servlet的定义

  <servlet>
      <servlet-name>servletTest2</servlet-name>
      <servlet-class>servletTest2</servlet-class>
  </servlet>
  
  <servlet-mapping>
      <servlet-name>servletTest2</servlet-name>
      <url-pattern>/test</url-pattern>
  </servlet-mapping>

例如项目名是test，运行tomcat，打开浏览器访问地址：http://127.0.0.1:8080/test/test

显示如下：

posted @ 2017-06-29 20:52  门罗的魔术师  阅读(8616)  评论(0)  编辑  收藏  举报