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jqueryjquery-uijquery-ui-draggablejquery-ui-droppable

jquery drag, drop and clone, find dropped position of element


Here, is my fiddle for jquery drag, drop and clone feature.

Problem:

My problem is: when ever I drop element, it is showing:

position: {top: 0, left: 0}

only for draggable, clone and droppable element.

I have also written code for finding position using only draggable function and that is working fine. I want this behavior in draggable, droppable with clone feature

Please visit JSFiddle

fiddle

full-screen output


Solution

  • Finally, resolved the problem. The problem was, I was using ui.draggable.position(); for storing dropped position to database, which was wrong.

    The actual position we need to store is:

    // position of the draggable minus position of the droppable
    // relative to the document
    leftPosition  = ui.offset.left - $(this).offset().left;
    topPosition   = ui.offset.top - $(this).offset().top;
    

    Reference: How do I get the coordinate position after using jQuery drag and drop?

    Updated Jsfiddle with working example

    http://jsfiddle.net/przbadu/rkvdffe3/18/

    http://jsfiddle.net/przbadu/rkvdffe3/18/embedded/result/