Jquery .clone()

Question

I'm currently trying to clone with the following code:

var position = $(this).position();
var ptop = position.top;
var pleft = position.left;
$(this).click(function() {
                $(this).clone().css({
                    top: ptop,
                    left: pleft,
                    opacity: '0.55'
                })
}).appendTo(this);

I need the element to clone at the exact position than the brother element. Thats why I have:

    var position = $(this).position();
    var ptop = position.top;
    var pleft = position.left;

For getting the position. But I also what the clone to have a lighter opacity.

Please be more specific. Your question is? The close is not working? What is happening wrong? — , Nov 23 '11 at 19:39
I assume it isn't work or you wouldn't be posting it here, but please be specific, what error do you get or what exactly is the problem? Maybe an example of the problem on jsFiddle.net would help you get a good answer. — Matt Cofer, Nov 23 '11 at 19:40
Sorry, Nothing is happening, the clone element is not being appended. — Shawn31313, Nov 23 '11 at 19:42

score 6 · Accepted Answer · answered Nov 23 '11 at 19:40

You've got your parentheses messed up. Your call to "appendTo()" is being applied after the establishment of the "click" handler.

$(this).click(function() {
                $(this).clone().css({
                    top: ptop,
                    left: pleft,
                    opacity: '0.55'
                }).appendTo(this);
});

You need it inside the "click" handler, and it needs to be called on the ".clone()" return value.

Jquery .clone()

1 Answers1