Question

UPDATED:

Using javascript or jQuery, how can I convert a number into it’s different variations:

eg:

1000000
to…

1,000,000 or 1000K

OR

1000
to…

1,000 or 1K

OR

1934 and 1234
to…

1,934 or -2K (under 2000 but over 1500)

or

1,234 or 1k+  (over 1000 but under 1500)

Can this is done in a function?

Hope this make sense.

C

Answer 1

You can add methods to Number.prototype, so for example:

Number.prototype.addCommas = function () {
    var intPart = Math.round(this).toString();
    var decimalPart = (this - Math.round(this)).toString();
    // Remove the "0." if it exists
    if (decimalPart.length > 2) {
        decimalPart = decimalPart.substring(2);
    } else {
        // Otherwise remove it altogether
        decimalPart = '';
    }
    // Work through the digits three at a time
    var i = intPart.length - 3;
    while (i > 0) {
        intPart = intPart.substring(0, i) + ',' + intPart.substring(i);
        i = i - 3;
    }
    return intPart + decimalPart;
};

Now you can call this as var num = 1000; num.addCommas() and it will return "1,000". That’s just an example, but you’ll find that all the functions create will involve converting the numbers to strings early in the process then processing and returning the strings. (The separating integer and decimal part will probably be particularly useful so you might want to refactor that out into its own method.) Hopefully this is enough to get you started.

Edit: Here’s how to do the K thing… this one’s a bit simpler:

Number.prototype.k = function () {
    // We don't want any thousands processing if the number is less than 1000.
    if (this < 1000) {
        // edit 2 May 2013: make sure it's a string for consistency
        return this.toString();
    }
    // Round to 100s first so that we can get the decimal point in there
    // then divide by 10 for thousands
    var thousands = Math.round(this / 100) / 10;
    // Now convert it to a string and add the k
    return thousands.toString() + 'K';
};

Call this in the same way: var num = 2000; num.k()