jQuery draggable element jumps back to its initial position

[narek]

Hello there .
I have some elements on my page that i have apllied jQuery's draggable function to them so I can drag them . Here is my question .
Is there a function which will return the draggable object to its initial position(the position where it was when the page loaded ) when an event occurs (for example I draggeed it and put it in a droppable element ) ?

Please I really need your help!!!
Thank you in advance!!

[SB65]

Not quite clear what you're trying to do, but if you specify

helper:'clone'

as an option for .draggable() then the dragged element will remain where it is and a clone of the element will be dragged - so the original element does not move. Is that what you mean? So this would allow you to build, for example, a shopping cart where you can drag items to the cart but they remain available in the shop for multiple purchases.

Otherwise, the revert option will cause the dragged element to return to it's original position when dragging stops.

[narek]

SB65 your post was really helpfull thank you.
But i miss only one thing here. After i drag the clone and dropp in the appropriate "drop element" how can i make it to disapprear(I mean hide or do watever to make the dragged element to disappear from its initial place ) ? And also how to make the "dragged" element be on the dropped element(since in case of clone it disappears) ?

[SB65]

To do something when the element is dropped it would be usual to write a function against the drop event:

$('#droppable').droppable({
	drop:function(event,ui){
               .... do things ......
	}
});

So this code could add something into the droppable element or whatever as required.

[narek]

SB65 I found a solution to my problem

$("#draggable").draggable({revert:true});

However thank you very much!!!