I have a link which I am going to use as notification when a user has some new notification I am going to notify the user by showing a tooltip(twitter bootstrap tooltip). Wh
You can add a variable to trigger off the mouseleave event to re-show the tooltip, and then as you said in your comment, just destroy the tooltip when clicked, so it doesn't show when you mouseover again:
var clickedNotify = false;
$('p a').tooltip({placement: 'bottom'}).tooltip('show');
$('p a').mouseleave(function() { if (!clickedNotify) { $('p a').tooltip({placement: 'bottom'}).tooltip('show'); } });
$('p a').click(function() { clickedNotify = true; $(this).tooltip('destroy'); });
This way, the tooltip is always shown, even after a mouseleave, until the link is clicked. After the link is clicked, the tooltip is destroyed, and still won't generate javascript errors on the page on mouseleave.
Here is the solution http://jsfiddle.net/testtracker/QsYPv/8/
Added the option "trigger"
$('p a').tooltip({placement: 'bottom',trigger: 'manual'}).tooltip('show');
then, with this line
$('p a').on('click',function(){$(this).tooltip('destroy');});
destroy tooltip on click.