5

So I do like this when user click a button:

var popup = $popup.create(url, features, options);
popup.open();

The problem is that every time the user clicks, a new popup is opened. I tried popup.isOpen() but is always false.

7

isOpen() is the correct method to call. It will return true any time in between the open() method being called and the close() method being called.

Maybe you are creating a new popup every time and that's why it is always false? You should keep the variable around (usually in this.properties) and check for null and isOpen() before you create a new popup.

Something like this:

var c = this.properties.controls;
if (c.popup && c.popup.isOpen())
{
   // Already open - do something (e.g. send new parameters to it)
   return;
} 

c.popup = $popup.create(url, features, options);
c.popup.open();
1
  • 3
    Only to add that the possible action to do when the popup is already open is c.popup.focus(); to bring it to the front view – Raúl Escudero Jan 22 '16 at 12:22

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