How to put this tooltip into a conditional in javascript?

Question

i have a problem when attemp put this Bootstrap "tooltip" into a conditional, the problem is that not show it. I show you my code:

$(document).ready(function () {
$("._diifem_").prop("disabled", true);
$('#wall').on('keyup', function () {
    var textarea_value = $("#wall").val();
    if (textarea_value.length > 5) {
        $('._diifem_').prop({
            disabled: false,
        });
    } else {
        $('._diifem_').prop({
            disabled: true
        }).tooltip({
            title: 'Hola',
            trigger: "click"
        });
    }
  })
});

The class "._diiefem" = Button that will be hidden. The ID "#wall" = textarea

what am I doing wrong? :(

Thanks everyone :3

Community · Accepted Answer · 2017-05-23 12:10:36Z

0

Look at the accepted answer of this question. Disabled buttons don't fire of any DOM events, I suspect that's why you're experiencing issues.

Check out some of the other answers to that question for possible workarounds. I think you'd probably be best off by faking a disabled button using the .disabled css class in Bootstrap.

edited May 23, 2017 at 12:10

CommunityBot

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answered May 6, 2017 at 17:52

bmceldowney

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1 Comment

Alex Parra Over a year ago

The problem with the .disabled class is that the button "status" is not updated :(

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