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Home/ Questions/Q 8997247
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T23:51:45+00:00 2026-06-15T23:51:45+00:00

I have two elements and will get strings inside. (and i use .each` function)

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I have two elements and will get strings inside. (and i use .each` function)

The problem is that the second array (after got string by .each), is replace the first one.

Sorry, if you don’t understand, but try to look below…

$('div').each(function () {
    var data = [];
    $('li', this).each(function () {
        data.push($(this).text());
    });
    var data_length = data.length;
    $(this).children("code").html(data + "");
    $("code").click(function () {
        data.move(data_length - 1, 0);
        $(this).html(data + "");
    });
});

Array.prototype.move = function (old_index, new_index) {
    if (new_index >= this.length) {
        var k = new_index - this.length;
        while ((k--) + 1) {
            this.push(undefined);
        }
    }
    this.splice(new_index, 0, this.splice(old_index, 1)[0]);
    return this; // for testing purposes
};

Demo: http://jsfiddle.net/kdpN7/

What did I do wrong?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-15T23:51:46+00:00Added an answer on June 15, 2026 at 11:51 pm

    For the same reason you do $(this).children('code') you should also bind your click event with a scope.

    The problem is, you’re iterating over 2 divs (your each) which means you’re binding $('code') twice. The first time code is bound to click, it binds with the first data array (the 1’s) and then it is bound a second time with (the 2’s). So it IS first doing your click code for the 1s and then immediately running it for the 2s, thus overwriting. Change to $(this).find("code") (or children) and it works as expected.

    http://jsfiddle.net/kdpN7/1/

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