Get string to array with multiple .each function

Question

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?

Your .move(l - 1, 0) can be written as a.unshift(a.pop()) if I'm not mistaken. — Ja͢ck
– Ja͢ck, Commented Dec 6, 2012 at 15:15

Eli Gassert · Accepted Answer · 2012-12-06 15:08:49Z

3

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/

answered Dec 6, 2012 at 15:08

Eli Gassert

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Comments

sbeliv01 · Accepted Answer · 2012-12-06 15:09:46Z

1

On this line:

$("code").click(function () { ...

This is telling to update all code with that information. You need to change it so it's specific to each div:

$(this).find("code").click(function () { ...

Updated fiddle: http://jsfiddle.net/kdpN7/2/

answered Dec 6, 2012 at 15:09

sbeliv01

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