i found problem like mine here, but no one topic help me, so i decide to write my own.
Problem : i need to 'mouseover' twice to see tip. And i know this is because i call tooltip on second mouseover, but i have no idea, how to solve my problem. I will be very thankful for any suggestion or help!
jquery :
$(document).ready(function() {
$('.playerTip').mouseover(function(){
var playerID = $(this).attr('rel');
$(this).tooltipster({
contentAsHTML: true,
content: 'Loading...',
functionBefore: function(origin, continueTooltip){
continueTooltip();
if(origin.data('ajax') !=='cached'){
$.ajax({
type: 'POST',
url: 'http://localhost/Statystyki/index.php/home/tooltip',
data: "player="+ playerID,
success: function(data) {
origin.tooltipster('content', data).data('ajax', 'cached');
}
});
}
}
});
});
});
and html:
<a href="#" class="playerTip" rel="426">XYZ</a>
<a href="#" class="playerTip" rel="174">DFV</a>
<a href="#" class="playerTip" rel="184">DRT</a>
etc
@solved well i solve my problem ;)
if anybody need solution:
$(document).ready(function() {
var playerID
$('.playerTip').mouseover(function(){
playerID = $(this).attr('rel');
});
$('.playerTip').tooltipster({
contentAsHTML: true,
content: 'Ładuje...',
functionBefore: function(origin, continueTooltip){
continueTooltip();
if(origin.data('ajax') !=='cached'){
$.ajax({
type: 'POST',
url: 'http://localhost/Statystyki/index.php/home/tooltip',
data: "player="+ playerID,
success: function(data) {
origin.tooltipster('content', data).data('ajax', 'cached');
}
});
}
}
});
});