I'm trying to achieve the following, can't figure out how though. I have several tooltips on a page, but I want each one to look different (depending on severity for instance). Here is what I have so far:
<div id="container">
<a href="#" title="Tooltip 1" data-severity="warning">Tooltip 1 (Warning)</a>
<a href="#" title="Tooltip 2" data-severity="danger">Tooltip 2 (Danger)</a>
</div>
<script type="text/javascript">
// init tooltip
ArrowToolTip.init('container');
</script>
/*JAVASCRIPT*/
var ArrowToolTip = {
init: function (parentId) {
jQuery('#' + parentId).tooltip(
{
content: function () {
var element = jQuery(this);
return element.attr('title');
},
create: function () {
var element = jQuery(this);
var severity = element.data('severity');
if (!severity) {
severity = 'default';
}
element.tooltip('option', 'tooltipClass', severity);
},
open: function(event, ui) {
var element = jQuery(this);
var severity = element.data('severity');
ui.tooltip({
tooltipClass: severity
});
//ui.tooltip.tooltipClass(severity);
console.log(event);
console.log(ui);
}
});
}
}
Whatever I have in the create: works, but it's only called on page load, so that doesn't do me any good. So now I'm trying to move this to the open:
event. But that doesn't seem to work at all. The jQuery(this)
doesn't return the hovered element, it returns the container. How can I get the correct element to select the data attribute? And how can I then change the CSS class? Does anyone know how this could be achieved?
I didn't get your logic very well, but here's how I achieve different classes depending on data-severity
. Instead of applying .tooltip()
to all elements, iterate through them and set the tooltipClass
as $(this).data('severity')
. The content function allows to add a class too.
$('a.tips').each(function () {
var severity = $(this).data('severity');
$(this).tooltip({
content: function () {
return '<em>' + $(this).attr('title') + '</em>';
},
tooltipClass: severity,
show: "slideDown",
open: function (event, ui) {
ui.tooltip.hover(function () {
$(this).fadeTo("slow", 0.5);
});
}
});
});
.ui-tooltip {
padding: 10px 20px;
border-radius: 20px;
font: bold 14px"Helvetica Neue", Sans-Serif;
text-transform: uppercase;
box-shadow: 0 0 7px black;
}
.ui-tooltip.warning {
color: #E3E8F7;
background: #D94AA4;
}
.ui-tooltip.danger {
color: #212942;
background: #CABA75;
}
a.tips { float:left; margin: 0 10px; text-decoration:none; font: Sans-Serif; }
body { background-color:#eee; padding: 30px; }
<link href="//cdnjs.cloudflare.com/ajax/libs/jqueryui/1.10.4/css/jquery.ui.tooltip.min.css" rel="stylesheet"/>
<script src="//ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script src="//cdnjs.cloudflare.com/ajax/libs/jqueryui/1.10.4/jquery-ui.min.js"></script>
<div id="container">
<a href="#" title="Tooltip 1" class="tips" data-severity="warning">(Warning)</a>
<a href="#" title="Tooltip 2" class="tips" data-severity="danger">(Danger)</a>
</div>
i am not clear about your quection. But i hope this will help you sometimes....
I am using tooltipClass name as "tooltipclass" in my jquery.
$(document).tooltip({effect: "blind", duration: 1000, tooltipClass: 'tooltipclass'});
And using that class name you can give any styling to the tool tip.
.tooltipclass{
font-size: 12px;
border:1px solid #e74c3c;
}
so try using different tooltipClass names.........
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