I am using jQuery draggable/droppable to enable elements in a grid to be swapped. For example, when element 1 is dropped onto element 2 they change places.
There are two problems:
After a successful drag-drop the clone still floats back to its original position. How can I prevent this? I only want it to float back if it's not dropped onto a valid target.
Ideally I'd like to set helper: original
as well, but when I tried
this I couldn't work out how to correctly calculate the swapped
positions in the drop
function.
I've put the code on jsFiddle, and below for reference:
<div id="room"></div>
<script>
$(function() {
//Set up seats - four rows of eight
for (var row = 0; row < 4; row++) {
for (var col = 0; col < 8; col++) {
$('#room').append('<div class="seat" style="top: ' + (row * 45 + 15) + 'px; left: ' + (col * 117 + 18) + 'px;">' + (row * 4 + col + 1) + '</div>');
}
}
//Swap function from http://blog.pengoworks.com/index.cfm/2008/9/24/A-quick-and-dirty-swap-method-for-jQuery
jQuery.fn.swap = function(b) {
b = jQuery(b)[0];
var a = this[0];
var t = a.parentNode.insertBefore(document.createTextNode(''), a);
b.parentNode.insertBefore(a, b);
t.parentNode.insertBefore(b, t);
t.parentNode.removeChild(t);
return this;
};
$(".seat").draggable({
revert: true,
helper: "clone"
});
$(".seat").droppable({
accept: ".seat",
hoverClass: "ui-state-hover",
drop: function(event, ui) {
var draggable = ui.draggable,
droppable = $(this),
dragPos = draggable.position(),
dropPos = droppable.position();
draggable.css({
left: dropPos.left + 'px',
top: dropPos.top + 'px'
});
droppable.css({
left: dragPos.left + 'px',
top: dragPos.top + 'px'
});
draggable.swap(droppable);
}
});
});
</script>
I resolved this with $(ui.helper).hide();
in the drop
function.