Hide bootstrap tooltip instead of disposing

Question

Is there any way we can hide bootstrap tool-tips instead of destroying them.

I am using server side data in tool-tips and I don't want to load data every time a tool-tip is initiated.

$('.selector').popover({
  animation: true,
  trigger: 'click',
  title: 'Notifications',
  content: 'No new notificaitons',
  placement: 'right',
  container: 'body',
  html: true,
})

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script src="https://stackpath.bootstrapcdn.com/bootstrap/4.1.1/js/bootstrap.bundle.min.js" integrity="sha384-u/bQvRA/1bobcXlcEYpsEdFVK/vJs3+T+nXLsBYJthmdBuavHvAW6UsmqO2Gd/F9" crossorigin="anonymous"></script>

<span class="selector">selector</span>

An alternative is to allow the tooltip to be destroyed but cache the results of whatever your call to the server is. — TKoL, Jun 05 '18 at 09:40

B3ND3L · Answer 1 · 2018-06-05T09:55:12.210

0

You don't have to hide your tooltip, you have an other way.

You can load your server side data in a global object and use this object when you open your tooltip.

In this case, your data is load only one times, and you don't have to change the tooltip.

Example :

var myObject = {};

$.get(PHP_PAGE, DATAS).done(function(data){
     //HERE GET YOUR DATA LIKE
     myObject = data;
});

...


$('.selector').popover({
  animation: true,
  trigger: 'click',
  title: myObject.title,
  content: myObject.contentMessage,
  placement: 'right',
  container: 'body',
  html: true,
})

edited Jun 05 '18 at 09:55

answered Jun 05 '18 at 09:48

B3ND3L

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Actually I am using a plugin which has to be initiated on any element, and that plugin fetches the data from the server. So I can not store data in any variable. . – hussain nayani Jun 05 '18 at 11:02

Hide bootstrap tooltip instead of disposing

1 Answers1