Reputation: 39
I have multiple fadeouts happening at the same time on various Ids, Is there any way of setting this up in one line as example:
This:
$("#bottle").on('click', function() {
$("#container_inner01").fadeIn(1100);
$("#container_inner02").fadeOut(1100);
$("#container_inner03").fadeOut(1100);
$("#container_inner04").fadeOut(1100);
$("#container02").fadeOut(1100);
})
turned it to this:
$("#bottle").on('click', function() {
$("#container_inner01").fadeIn(1100);
$("#container_inner02,#container_inner02,#container_inner03,#container_inner04,#container02").fadeOut(1100);
})
I am sure its possible and its some syntax error I am doing.
Upvotes: -1
Views: 82
Reputation: 21565
Doing what you have in the second script:
$("#bottle").on('click', function() {
$("#container_inner01").fadeIn(1100);
$("#container_inner02,#container_inner02,#container_inner03,#container_inner04,#container02").fadeOut(1100);
});
Will work, although you can sum this up to this:
$("#bottle").on('click', function() {
$("#container_inner01").fadeIn(1100);
$("#container_inner02,#container_inner03,#container_inner04,#container02").fadeOut(1100);
});
Note that you have #container_inner02,#container_inner02
in you script repeating. Another way to do this is to add a class to all of them and using that class affects each item:
$("#bottle").on('click', function() {
$("#container_inner01").fadeIn(1100);
$(".myClassName").fadeOut(1100);
});
Upvotes: 0
Reputation: 664
Try this
(function($){
$("#bottle").on('click', function() {
$('#container_inner01').fadeIn(1100);
$('#container_inner02,#container_inner03,#container_inner04,#container02').fadeOut(1100);
});
})(jQuery);
Upvotes: 0
Reputation: 2667
Why not give your elements a class
and execute the script that way?
$("#bottle").on('click', function () {
$("#container_inner01").fadeIn(1100);
$('.test').fadeOut(1100);
});
Upvotes: 2