When I make a draggable clone and drop it in a droppable I cannot drag it again

user_104833 0
Viewed 97081

When I make a draggable clone and drop it in a droppable I cannot drag it again. How do I do that? Secondly I can only figure out how to us .append to add the clone to the droppable. But then it snaps to the top-left corner after any existing element and not the drop position.

$(document).ready(function() {
    $("#container").droppable({
        drop: function(event, ui) {
            $(this).append($(ui.draggable).clone());
        }
    });
    $(".product").draggable({
        helper: 'clone'
    });
});
</script>

<div id="container">
</div>
<div id="products">
    <img id="productid_1" src="images/pic1.jpg" class="product" alt="" title="" />
    <img id="productid_2" src="images/pic2.jpg" class="product" alt="" title="" />
    <img id="productid_3" src="images/pic3.jpg" class="product" alt="" title="" />
</div>
jquery-ui-droppable jquery-ui-draggable drag-and-drop clone jquery
4 Answers

One way to do it is:

$(document).ready(function() {
    $("#container").droppable({
        accept: '.product',
        drop: function(event, ui) {
            $(this).append($("ui.draggable").clone());
            $("#container .product").addClass("item");
            $(".item").removeClass("ui-draggable product");
            $(".item").draggable({
                containment: 'parent',
                grid: [150,150]
            });
        }
    });
    $(".product").draggable({
        helper: 'clone'
    });
});

But I'm not sure if it is nice and clean coding.

user_1048330

For those trying to reposition the dropped item. Take a look here.

Jquery drag /drop and clone.

I actually had to use code that looks like

$(item).css('position', 'absolute');
$(item).css('top', ui.position.top - $(this).position().top);
$(item).css('left', ui.position.left - $(this).position().left);

to do it.

user_1273880
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