打开网站但仅是第一次时打开弹出窗口. [英] open popup window when website is open but only first time.

问题描述

朋友们.
我有一个要求，当用户打开网站时打开弹出消息框，但它应该仅是第一次打开.假设用户关闭窗口并再次打开网站，则应该再次显示它.


问候
Imrankhan

Hi friends.
I have one requirement to open popup message box when user open website but it should be open first time only. Suppose user close window and open website again then it should be display again.


Regard
Imrankhan

推荐答案

嘿，您可以使用cookie解决问题.跟踪脚本可以使用Cookie来确定访问者是否曾经来过这里，并且仅在他们第一次访问该页面时打开一个新窗口.下次他们回来时，脚本将读取cookie，将其标识为重复访问者，而不会再次打开窗口.

Hey you can solve the problem using cookies. Using cookies, follwing script can determine if the visitor has been here before, and only open a new window on their first visit to the page. The next time they come back, the script will read the cookie, identify them as a repeat visitor, and not open the window again.

<pre lang="cs">var expDays = 1; // number of days the cookie should last

var page = "only-popup-once.html";
var windowprops = "width=300,height=200,location=no,toolbar=no,menubar=no,scrollbars=no,resizable=yes";

function GetCookie (name) {
  var arg = name + "=";
  var alen = arg.length;
  var clen = document.cookie.length;
  var i = 0;
  while (i < clen) {
    var j = i + alen;
    if (document.cookie.substring(i, j) == arg)
    return getCookieVal (j);
    i = document.cookie.indexOf(" ", i) + 1;
    if (i == 0) break;
  }
  return null;
}

function SetCookie (name, value) {
  var argv = SetCookie.arguments;
  var argc = SetCookie.arguments.length;
  var expires = (argc > 2) ? argv[2] : null;
  var path = (argc > 3) ? argv[3] : null;
  var domain = (argc > 4) ? argv[4] : null;
  var secure = (argc > 5) ? argv[5] : false;
  document.cookie = name + "=" + escape (value) +
    ((expires == null) ? "" : ("; expires=" + expires.toGMTString())) +
    ((path == null) ? "" : ("; path=" + path)) +
    ((domain == null) ? "" : ("; domain=" + domain)) +
    ((secure == true) ? "; secure" : "");
}

function DeleteCookie (name) {
  var exp = new Date();
  exp.setTime (exp.getTime() - 1);
  var cval = GetCookie (name);
  document.cookie = name + "=" + cval + "; expires=" + exp.toGMTString();
}

var exp = new Date();
exp.setTime(exp.getTime() + (expDays*24*60*60*1000));

function amt(){
  var count = GetCookie(''count'')
  if(count == null) {
    SetCookie(''count'',''1'')
    return 1
  } else {
    var newcount = parseInt(count) + 1;
    DeleteCookie(''count'')
    SetCookie(''count'',newcount,exp)
    return count
  }
}

function getCookieVal(offset) {
  var endstr = document.cookie.indexOf (";", offset);
  if (endstr == -1)
  endstr = document.cookie.length;
  return unescape(document.cookie.substring(offset, endstr));
}

function checkCount() {
  var count = GetCookie(''count'');
  if (count == null) {
    count=1;
    SetCookie(''count'', count, exp);
    window.open(page, "", windowprops);
  } else {
    count++;
    SetCookie(''count'', count, exp);
  }
}

window.onload=checkCount;

请尝试以下代码:

Hi,

Please try the following code:

protected void Page_Load(object sender, EventArgs e)
        {
            if (!IsPostBack)
            {
                if (Session["PopupDisplayed"] == null)
                {
                    ClientScript.RegisterStartupScript(GetType(), "DisplayPopup", "window.open('popupurlhere');", true);
                    Session["PopupDisplayed"] = true;
                }
            }
        }

问候，
贾米尔(Jamil)

Regards,
Jamil

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