This is a follow up to my previous question.
I need to place Bootstrap tooltips dynamically. To do so, I need to know the position of the element which triggers the tooltip (works fine) AND the width of the tooltip which is supposed to show up:
Edit for clarification: My tooltips are generated dynamically and then inserted into the content. If they are, for example, too close to the left edge of the screen I need to place them 'right' instead of 'top'. Now I want to get the width of the tooltip to only place it 'right' when it would actually go out of the screen.
$('body').tooltip({
delay: { show: 300, hide: 0 },
selector: '[rel=tooltip]:not([disabled])',
placement: function(tip, element) {
var offsetLeft = $(element).offset().left;
var offsetRight = ( $(window).width() - ( offsetLeft + $(element).outerWidth() ) );
// need help here: how to get the width of the current tooltip?
// currently returns "0" for all elements
var tooltipWidth = $(tip).outerWidth();
console.log(tooltipWidth);
if (offsetLeft < 220) {
return 'right';
}
if (offsetRight < 220) {
return 'left';
}
else {
return 'top';
}
}
});
Thanks in advance.
From a logically point of view it is impossible :) Think about it - the tooltip calls the placement function to get its position, and then it inserts the tip to the page and style it.
However, you can create a dummy-tip with the same features as the soon-to-come tip, and by that get the width. Like this :
$('body').tooltip({
delay: { show: 300, hide: 0 },
placement: function(a, element) {
//title is by tooltip moved to data-original-title
var title=$(element).attr('data-original-title');
//create dummy, a div with the same features as the tooltïp
var dummy=$('<div class="tooltip">'+title+'</div>').appendTo('body');
var width=$(dummy).width();
dummy.remove();
//now you have width
//..
//..
var position = $(element).position();
if (position.left > 515) {
return "left";
}
if (position.left < 515) {
return "right";
}
if (position.top < 110){
return "bottom";
}
return "top";
},
selector: '[rel=tooltip]:not([disabled])'
});
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