防止页面上的setTimeout函数重新加载

Question

I am running on Bootstrap 3.3.6. I have a very simple setTimeout function that fires a popup window after 10 sec.

setTimeout(function(){
        $('#myModal').modal('show');
        }, 10000);

I added this script at the end of my homepage. This code works perfectly and the popup (#myModal) opens up just fine.

Problem is, if I visit another page on my site and then hit the back button, or click home on the navbar, the pop up fires again.

I found an answer on here about using localStorage and running it as an if/else. Basically make it so that you will only see the popup the first time you visit that page. But I am very new to javascript and I am having a hard time getting it right. Can someone help me out here?

Answer 1

Well, I suppose one way to do it might be to create a cookie, and then check for it. Perhaps you could create a cookie like this in the script for your home page:

 window.onload = function() { if (!document.cookie){ setTimeout(function(){ $('#myModal').modal('show'); }, 10000); document.cookie = "cookie=yes; path=/"; } } 

If I've done everything properly, what this should do first is check to see if there isn't a cookie. If there isn't a cookie, your modal alert goes off and a cookie with a name of "cookie" and a value of "yes" gets created. Since we are not specifying a max age for this cookie, it should expire when the user leaves the site (I think). Also, since we set a general path, this cookie should persist across every page.

Answer 2

I would use a simple approach like this:

 // load flag from localStorage var alreadyDisplayed = localStorage.getItem('alreadyDisplayed'); // if not found or found but 'false' - display modal (alert in this case) if (!alreadyDisplayed || alreadyDisplayed == 'false') { setTimeout(function() { alert('show'); // store to localStorage the fact that we displayed the modal localStorage.setItem('alreadyDisplayed', 'true'); }, 5 * 1000); // in ms, so 5*1000 = 5 secs }