当用户离开我的网站时显示弹出窗口

Question

I want that user should be given a popup whenever he/she navigates away from my website(not just the webpage, the entire site).

I have implemented a method but it pop-up every time on navigating away from webpage.

window.onbeforeunload = function (evt) {


  var message = 'Are you sure you want to navigate away from myWEbSite?';
  if (typeof evt == 'undefined') {
    evt = window.event;

  }
  if (evt ) {
    evt.returnValue = message;
  }

  return message;

 }

Answer 1

You cannot get the new target url, since this is a protected information (see How can i get the destination url in javascript onbeforeunload event? )

So, you are not able to see, if the visitor leaves the website.

Alternative: you could listen all clicks on hyperlinks which refer to another webpage and implement a functionality which catches external links.

Answer 2

You can defile your own JQuery selectors, like this:

$.expr[':'].external = function(obj){
   return !obj.href.match(/^mailto\:/) && (obj.hostname != location.hostname);
};
$.expr[':'].internal = function(obj){
   return obj.hostname == location.hostname;
};

And then you can do the following:

$(function() {
    var unloadMessage = function() {
        return "Are you sure you want to navigate away from myWEbSite?";
    };

    $('a:internal').click(function() { 
        window.onbeforeunload = null;
    });
    $('form').submit(function() { 
        window.onbeforeunload = null;
    });

    $('a:external').click(function() { 
        window.onbeforeunload = unloadMessage;
    });

    window.onbeforeunload = unloadMessage;
});

In your specific case you just need to set window.onbeforeunload == null whenever a navigational link is clicked or an internal form is submitted.