1.建立web工程
2.实现servlet
package lin;
import javax.servlet.*;
import javax.servlet.http.*;
import java.io.*;
public class servlet extends GenericServlet{
public void service(ServletRequest request,ServletResponse response)throws IOException , ServletException{
response.setCharacterEncoding("GBK");
PrintWriter out = response.getWriter();
out.println("");
out.println("
");out.println("
out.println("");
out.println("
");out.println("
out.println("
out.println("
out.println("");
out.println("");
out.close();
}
}
3.在工程的web.xml配置servlet访问url
4.发布工程到tomcat
5.启动tomcat
6.locatlhost:8080/工程名/servlet路径
例如我给出来的例子就是运行这个:localhost:8888/SampleServlet/servlet
补充:
1、在xm中有多个
2. MyEclipse 8.5 会自动配置web.xml文件,只需注意查看xml文件中的servlet访问路径。
切换到sourse默认打开的是design视图