Remove cloned image after adding another image wit

问题:

I have a nice solution from my previous question that successfully clones images after being dropped.

Here is the code:

$(function() {
var makeDraggable = function(element) {
    element.draggable({ 
        revert: "invalid",
        appendTo: "#droppable",
        helper: "clone" 
    });
}
$( "#droppable" ).droppable({
    activeClass: "ui-state-default",
    hoverClass: "ui-state-hover",
    drop: function(event, ui) {
        var newClone = $(ui.helper).clone();
        makeDraggable(newClone);
        $(this).after(newClone);
    }
});
// Initalise the program by making first draggable:
makeDraggable($(".draggable img"));

But the problem is I want to show only one image at a time in the targeted area. But currently all the dropped images are shown.

More specifically when user drops an image in the targeted area and later drags another image, the previous image should be removed from the dropped or targeted area and only the new image should be visible in the targeted area. See this demo: jsFiddle example

How do I solve this?

回答1:

Instead of simply emptying the droppable target area like the other answers do, I would add a class to the dropped items and then remove them upon a dragstart event of a new draggable. Also, it's nice to add a little fadeOut() event when a new draggable is selected. However, as Ishettyl pointed out, one must also fadeIn() the element again if the user decides not to drop the draggable. This can be done using a custom revert function (see below).

The result is then something like this:

In my opinion this looks more elegant and doesn't confuse the user whether more items are allowed.

Code

$(function() {
    $(".draggable img").draggable({ 
        revert: function (socketObj) {
            if (!socketObj) {
                if ($(this).hasClass("droppedItems")) {
                    $(this).fadeOut(function() {
                        $(this).remove();
                    });
                } else {
                    $(".droppedItems").fadeIn();
                    return true;   
                }
            }
        },
        helper: "clone",
        start: function(event, ui) {
            $(".droppedItems").fadeOut();
        }
    }); 
    $( "#droppable" ).droppable({
        activeClass: "ui-state-default",
        hoverClass: "ui-state-hover",
        drop: function(event, ui) {
            $(".droppedItems").remove();
            var newClone = $(ui.helper).clone();
            newClone.addClass("droppedItems");
            $(this).append(newClone);
        }
    });
});

DEMO

回答2:

In the drop method instead of using .after use .append() like this

drop: function(event, ui) {
  var newClone = $(ui.helper).clone();
  makeDraggable(newClone);
  $(this).children(':not(span)').remove(); // <--- this will remove all the elements which are not a span before appending
  $(this).append(newClone);
}

Also, instead of writing drop inside div#droppable use a span like this

<div id="droppable">
  <span>drop</span>
</div>

Here is a demo http://jsfiddle.net/dhirajbodicherla/e3pf6ays/14/

回答3:

You were adding the clones one after the other with the below code.

$(this).after(newClone);

All you need to do is: empty the droppable container and then add the new clones, like below:

drop: function(event, ui) {
    var newClone = $(ui.helper).clone();
    makeDraggable(newClone);
    $(this).empty().append(newClone);
}

Updated fiddle