i have here three div boxes which i included into a rotator script .
So if some one clicks on the right button it will shows the next div box. The problem that i have is that it shows me all 3 boxes. as u can see here. So they laying over each other.
How do i make it that it just shows me one and the other one by click on the button ?
$(document).ready(function(){
//===== Apps slider script =====
var angle = 0;
$('.slider .slide:odd').css({
"-webkit-transform": "rotateY(180deg)",
"-moz-transform": "rotateY(180deg)",
"-o-transform": "rotateY(180deg)",
"-ms-transform": "rotateY(180deg)",
"transform": "rotateY(180deg)"
/* "-moz-transform": "scaleX(-1)",
"-o-transform": "scaleX(-1)",
"-webkit-transform": "scaleX(-1)",
"-ms-transform": "scaleX(-1)",
"transform": "scaleX(-1)"*/
});
function sliderResize(){
if ($(window).width() <= 550) {
$('.appsblock .apps .slider').css("min-height", $(window).width() + 100);
}
}
$('.slider .navigation-right').click(function(){
if ($(this).parent().find('.active').is(':last-child') == false) {
angle = angle - 180;
var angledeg = 'rotateY(' + angle + 'deg)';
$(this).parent().find('.rotator').css({
"-webkit-transform": angledeg,
"-moz-transform": angledeg,
"-o-transform": angledeg,
"-ms-transform": angledeg,
"transform": angledeg
});
$(this).parent().find('.active').next().toggleClass('active');
$(this).parent().find('.active:first').toggleClass('active');
}
});
$('.slider .navigation-left').click(function(){
if ($(this).parent().find('.active').is(':first-child') == false) {
angle = angle + 180;
var angledeg = 'rotateY(' + angle + 'deg)';
$(this).parent().find('.rotator').css({
"-webkit-transform": angledeg,
"-moz-transform": angledeg,
"-o-transform": angledeg,
"-ms-transform": angledeg,
"transform": angledeg
});
$(this).parent().find('.active').prev().toggleClass('active');
$(this).parent().find('.active:last').toggleClass('active');
}
});
I put in line 540 a opacity:0, that was on my old side to and there it works perfactly. if you take the opacity on line 540 in the css part put you will see what I mean.