Bootstrap: tooltip/popover 'destroy' needs event

Created on 19 Jan 2015 · 8Comments · Source: twbs/bootstrap

The tooltip/popover 'destroy' method is implicitly async because it calls hide first, and so a sequence like $(selector).popover('destroy').popover({ ... }) does not reliably work.

I suggest that, like 'show' and 'hide', 'destroy' be given events 'destroy.bs.popover' and 'destroyed.bs.popover', so that a caller can know for sure that a 'destroy' is finished and can do whatever they need to do to the elements which depends on the tooltip/popover being cleared out.

feature js

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dondi

👍1

Most helpful comment

@fat but we need it for Bootstrap 3 also. :disappointed:

seyfer on 15 Oct 2018

👍2

All 8 comments

Seconded, Chain All the things!

SaymV on 19 Jan 2015

Agreed, chaining would be very useful.

bitcity on 20 Jan 2015

+1. Killed some hours while working with tooltips.

pavel-lebedinsky on 22 Jan 2015

👍1

Is this a bug or a feature @cvrebert? If the latter, shall we punt to v4?

mdo on 27 Mar 2015

I think the real problem here is that "destroy" was conflated with "hide", making it async… which is probably not the best.

At any rate, we are going to revisit teardown methods in v4, thanks for the input!

fat on 25 Apr 2015

@fat but we need it for Bootstrap 3 also. :disappointed:

seyfer on 15 Oct 2018

👍2

@fat was it fixed in bootstrap v3.4.0? seems like not...

Expected behavior:

$('#destroy-confirmation')
      .tooltip({placement: 'top'})
      .tooltip('destroy');

  $('#destroy-confirmation-re')
      .tooltip({placement: 'top'})
      .tooltip('destroy')
      .tooltip({placement: 'right'});

1st case - doesn't show tooltip after destroy
2nd case - show on the right since it was re-initialized

seyfer on 13 Feb 2019

@seyfer did you find some workaround?