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XMLHttpServlet的open()方法有关问题

发布时间: 2013-07-08 14:13:00 作者: rapoo

XMLHttpServlet的open()方法问题,求助
本帖最后由 showbo 于 2013-07-04 18:10:42 编辑

//创建XMLHTTPREQUEST组件
xmlHttp=createXmlHttpRequest();
//设置回调函数
xmlHttp.onreadystatechange=processRequest;
//打开与服务器响应地址的连接
xmlHttp.open("get","CheckUserServlet?uname="+uname,true);/////
//发送请求
xmlHttp.send(null);



以下是CheckUserServlet代码

public class CheckUserServlet extends HttpServlet {

private static final long serialVersionUID = -2105660521556754174L;


public void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
request.setCharacterEncoding("utf-8");
String user=request.getParameter("uname");
UserModel um=new UserModel();
response.setCharacterEncoding("utf-8");
PrintWriter out=response.getWriter();
//返回信息
out.print(um.validate(user));
out.flush();
out.close();
response.setContentType("text/html");
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();
}
XMLHttpServlet
[解决办法]
没看出什么问题。注意状态转换函数定义了没有,还有就是get加时间戳防止ie缓存

xmlHttp.open("get","CheckUserServlet?uname="+uname+'&_dc='+new Date().getTime(),true);

ie8+有开发人员工具,自己f12调出来看什么错误

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