ajax请求servlet刷新html页面

1、java类

    package testsub;
    public class Sys {
        /**
        * 获取到一条消息
        **/
	public static String getMessage() {
		return "模拟,请求后端,返回一条消息";
	}
    }

2、servlet类

    package testsub;
    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 MyServlet extends HttpServlet{
	private static final long serialVersionUID = -7820821330469179359L;
	@Override
	public void init() throws ServletException {
		super.init();
	}
	@Override
	protected void doGet(HttpServletRequest req, HttpServletResponse resp)
			throws ServletException, IOException {
		req.setCharacterEncoding("utf-8");
		resp.setContentType("text/html;charset=utf-8");
                req.getParameter("selecttime");//获取ajax请求参数
		String str=Sys.getMessage;
		resp.getWriter().print(str);
	}
	@Override
	protected void doPost(HttpServletRequest req, HttpServletResponse resp)
			throws ServletException, IOException {
		doGet(req, resp);
	}
    }

3、servlet配置

    <servlet>
 	<servlet-name>myservlet</servlet-name>
 	<servlet-class>testsub.MyServlet</servlet-class>
    </servlet>
    <servlet-mapping>
 	<servlet-name>myservlet</servlet-name>
 	<url-pattern>*.do</url-pattern>
    </servlet-mapping>

4、html代码




测试请求后台数据成功否






posted @ 2017-04-28 11:07  goumingming  阅读(4160)  评论(0)    收藏  举报