How to run function when input is disabled or reenabled

Question

Is there some way to run a function when a input is disabled/enabled from another part of the code, for example when this code is run:

$('#myinput').prop('disabled', true);

I want another part of the code to be notified about this, but without making it dependent on the other part. Something similar to how "change" event works. But there's no "disabled" event...

Trigger that event to notify about update; – itzmukeshy7 Apr 11 '16 at 09:04 — itzmukeshy7, Apr 11 '16 at 09:04
By the way what you want to do in this way? – itzmukeshy7 Apr 11 '16 at 09:04 — itzmukeshy7, Apr 11 '16 at 09:04

score 0 · Answer 1 · edited May 23 '17 at 11:50

0

There is no event that will trigger when those properties are changed.

You can do this some other way.

Set hidden field for this enable disabled value like 1 / 0

$("#hiddenid").trigger("hiddenidchange");


$("#hiddenid").on("hiddenidchange", function () {

});

refer How do I trigger an onChange event for a hidden field?

edited May 23 '17 at 11:50

Community

1
1

answered Apr 11 '16 at 09:49

Ponmani Chinnaswamy

705
7
21

score 0 · Answer 2 · answered Apr 11 '16 at 09:51

0

You can call notification trigger in the next line with an if statement

document.getElementById("email").disabled=true;
if(document.getElementById("email").disabled == true)
 {
   alert("Notifying..");
   //call the trigger function
 }

answered Apr 11 '16 at 09:51

Rakesh Bk

112
15

How to run function when input is disabled or reenabled

2 Answers2