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Home/ Questions/Q 8755897
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T13:56:43+00:00 2026-06-13T13:56:43+00:00

Take the following example: $.fn.foo = function() { var $input = $(‘<input type=text/>’); var

  • 0

Take the following example:

$.fn.foo = function() {

    var $input = $('<input type="text"/>');
    var $button_one = $('<button>One</button>');
    var $button_two = $('<button>Two</button>');

    $button_one.click(function(){
        $input.val('hey');
    });

    $button_two.click(function(){
        $input.replaceWith( $input.val('').clone( true ) );
    });

    this.append( $input, $button_one, $button_two );


};

Check the demo: http://jsbin.com/ezexah/1/edit

Now click “one” and you should see “hey” in the input. Next click “two” and then click “one” again and it doesn’t work. Even using the true option in clone to copy all events it still does not work. Any ideas?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-13T13:56:44+00:00Added an answer on June 13, 2026 at 1:56 pm

    You did not update $input. It is still referring to the element you just removed, meaning $input.val('hey'); will update the value of the DOM element that is now detached from the DOM tree.
    .clone or .replaceWith do not automatically update existing references:

    var $clone = $input.val('').clone( true )
    $input.replaceWith($clone);
    $input = $clone;
    

    This has nothing to do with the true option btw, because you did not bind any event handlers to $input.

    DEMO

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