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I have the following function that automatically adds the commas to an American numerical expression which is some way of localisation. I need a minor change to this function.
if I have 160 * 54.08 = 8.652,8 as input to the function what can I do to show the output as 8.652,80 with two decimal places? Right now the function outputs as 8.652.8
function Numberformat(nStr) {
nStr += '';
x = nStr.split('.');
x1 = x[0];
x2 = x.length > 1 ? ',' + x[1] : '';
var rgx = /(\d+)(\d{3})/;
while (rgx.test(x1)) {
x1 = x1.replace(rgx, '$1' + '.' + '$2');
}
return x1 + x2;
}
function Numberformat(nStr) {
nStr += '';
x = nStr.split('.');
x1 = x[0];
x2 = x.length > 1 ? ',' + x[1] : ',0';//if there is no decimal
//force ",00"
x2 = x2.length < 3? x2 + "0" : x2;//if length of decimal + delimiter is
//less than 3 add an extra 0
var rgx = /(\d+)(\d{3})/;
while (rgx.test(x1)) {
x1 = x1.replace(rgx, '$1' + '.' + '$2');
}
return x1 + x2;
}
document.body.innerHTML = Numberformat(160 * 54.08);//8.652,80
document.body.innerHTML += '<br>' + Numberformat(8560);//8.560,00
document.body.innerHTML += '<br>' + Numberformat(8560.0);//8.560,00
document.body.innerHTML += '<br>' + Numberformat(8560.1);//8.560,10
document.body.innerHTML += '<br>' + Numberformat(8560.01);//8.560,01
document.body.innerHTML += '<br>' + Numberformat(8560.009);//8.560,009
Check this answer:https://stackoverflow.com/a/149099/3025534
You can use:
var profits=2489.8237
profits.toFixed(3) //returns 2489.824 (round up)
profits.toFixed(2) //returns 2489.82
profits.toFixed(7) //returns 2489.8237000 (padding)
Then you can add the sign of '$'.
If you require ',' for thousand you can use:
Number.prototype.formatMoney = function(c, d, t){
var n = this,
c = isNaN(c = Math.abs(c)) ? 2 : c,
d = d == undefined ? "." : d,
t = t == undefined ? "," : t,
s = n < 0 ? "-" : "",
i = parseInt(n = Math.abs(+n || 0).toFixed(c)) + "",
j = (j = i.length) > 3 ? j % 3 : 0;
return s + (j ? i.substr(0, j) + t : "") + i.substr(j).replace(/(\d{3})(?=\d)/g, "$1" + t) + (c ? d + Math.abs(n - i).toFixed(c).slice(2) : "");
};
And use it with:
(123456789.12345).formatMoney(2, '.', ',');
If you're always going to use '.' and ',', you can leave them off your method call, and the method will default them for you.
(123456789.12345).formatMoney(2);
If your culture has the two symbols flipped (i.e. Europeans), just paste over the following two lines in the formatMoney method:
d = d == undefined ? "," : d,
t = t == undefined ? "." : t,
I would do something like this
var n = '1.000,00';
if (wrongFormat(n)) {
n = n.replace(',', '#');
n = n.replace('.', ',');
n = n.replace('#', '.');
}
var val = 1*n;
//do something with val here
This question already has answers here:
How can I pad a value with leading zeros?
(76 answers)
Closed 4 years ago.
How can I modify this code to add a 0 before any digits lower than 10
$('#detect').html( toGeo(apX, screenX) + latT +', '+ toGeo(apY, screenY) + lonT );
function toGeo(d, max) {
var c = '';
var r = d/max * 180;
var deg = Math.floor(r);
c += deg + "° ";
r = (r - deg) * 60;
var min = Math.floor(r);
c += min + "′ ";
r = (r - min) * 60;
var sec = Math.floor(r);
c += sec + "″";
return c;
}
So the outpout would change from
4° 7′ 34″W, 168° 1′ 23″N
to
04° 07′ 34″W, 168° 01′ 23″N
Thanks for your time
You can always do
('0' + deg).slice(-2)
See slice():
You can also use negative numbers to select from the end of an array
Hence
('0' + 11).slice(-2) // '11'
('0' + 4).slice(-2) // '04'
For ease of access, you could of course extract it to a function, or even extend Number with it:
Number.prototype.pad = function(n) {
return new Array(n).join('0').slice((n || 2) * -1) + this;
}
Which will allow you to write:
c += deg.pad() + '° '; // "04° "
The above function pad accepts an argument specifying the length of the desired string. If no such argument is used, it defaults to 2. You could write:
deg.pad(4) // "0045"
Note the obvious drawback that the value of n cannot be higher than 11, as the string of 0's is currently just 10 characters long. This could of course be given a technical solution, but I did not want to introduce complexity in such a simple function. (Should you elect to, see alex's answer for an excellent approach to that).
Note also that you would not be able to write 2.pad(). It only works with variables. But then, if it's not a variable, you'll always know beforehand how many digits the number consists of.
Make a function that you can reuse:
function minTwoDigits(n) {
return (n < 10 ? '0' : '') + n;
}
Then use it in each part of the coordinates:
c += minTwoDigits(deg) + "° ";
and so on.
if(myNumber.toString().length < 2)
myNumber= "0"+myNumber;
or:
return (myNumber.toString().length < 2) ? "0"+myNumber : myNumber;
You can always do
('0' + deg).slice(-2)
If you use it very often, you may extend the object Number
Number.prototype.pad = function(n) {
if (n==undefined)
n = 2;
return (new Array(n).join('0') + this).slice(-n);
}
deg.pad(4) // "0045"
where you can set any pad size or leave the default 2.
You can write a generic function to do this...
var numberFormat = function(number, width) {
return new Array(+width + 1 - (number + '').length).join('0') + number;
}
jsFiddle.
That way, it's not a problem to deal with any arbitrarily width.
Hope, this help:
Number.prototype.zeroFill= function (n) {
var isNegative = this < 0;
var number = isNegative ? -1 * this : this;
for (var i = number.toString().length; i < n; i++) {
number = '0' + number;
}
return (isNegative ? '-' : '') + number;
}
Here is Genaric function for add any number of leading zeros for making any size of numeric string.
function add_zero(your_number, length) {
var num = '' + your_number;
while (num.length < length) {
num = '0' + num;
}
return num;
}
I was bored and playing around JSPerf trying to beat the currently selected answer prepending a zero no matter what and using slice(-2). It's a clever approach but the performance gets a lot worse as the string gets longer.
For numbers zero to ten (one and two character strings) I was able to beat by about ten percent, and the fastest approach was much better when dealing with longer strings by using charAt so it doesn't have to traverse the whole string.
This follow is not quit as simple as slice(-2) but is 86%-89% faster when used across mostly 3 digit numbers (3 character strings).
var prepended = ( 1 === string.length && string.charAt( 0 ) !== "0" ) ? '0' + string : string;
$('#detect').html( toGeo(apX, screenX) + latT +', '+ toGeo(apY, screenY) + lonT );
function toGeo(d, max) {
var c = '';
var r = d/max * 180;
var deg = Math.floor(r);
if(deg < 10) deg = '0' + deg;
c += deg + "° ";
r = (r - deg) * 60;
var min = Math.floor(r);
if(min < 10) min = '0' + min;
c += min + "′ ";
r = (r - min) * 60;
var sec = Math.floor(r);
if(sec < 10) sec = '0' + sec;
c += sec + "″";
return c;
}
A single regular expression replace should do it:
var stringWithSmallIntegers = "4° 7′ 34″W, 168° 1′ 23″N";
var paddedString = stringWithSmallIntegers.replace(
/\d+/g,
function pad(digits) {
return digits.length === 1 ? '0' + digits : digits;
});
alert(paddedString);
shows the expected output.
Good day,
Is there a Regex that I could use to prepend a 0 before any number that is below 10?
I am not looking for a date parsing library, ternary or if/else solutions. (hopefully)
var currentDate = new Date(),
stringDate = currentDate.getFullYear() + "-" + currentDate.getMonth() + "-" + currentDate.getDate() + " " + currentDate.getHours() + ":" + currentDate.getMinutes() + ":" + currentDate.getSeconds();
alert( stringDate ); //2011-10-17 10:3:7
I would like a RegExp that I could apply to stringDate to get 2011-10-17 10:03:07
Thank you very much!
Just add the leading 0 every time, then use slice(-2) to get the last two characters, like so:
('0' + currentDate.getHours()).slice(-2)
The following function will allow you to declare a minimum length for a number along or within a string and will pad it with zeros to make it the appropriate length.
var PrependZeros = function (str, len, seperator) {
if(typeof str === 'number' || Number(str)){
str = str.toString();
return (len - str.length > 0) ? new Array(len + 1 - str.length).join('0') + str: str;
}
else{
for(var i = 0,spl = str.split(seperator || ' '); i < spl.length; spl[i] = (Number(spl[i])&& spl[i].length < len)?PrependZeros(spl[i],len):spl[i],str = (i == spl.length -1)?spl.join(seperator || ' '):str,i++);
return str;
}
};
For those wanting a less cryptic version
var PrependZeros = function (str, len, seperator) {
if (typeof str === 'number' || Number(str)) {
str = str.toString();
return (len - str.length > 0) ? new Array(len + 1 - str.length).join('0') + str : str;
}
else {
var spl = str.split(seperator || ' ')
for (var i = 0 ; i < spl.length; i++) {
if (Number(spl[i]) && spl[i].length < len) {
spl[i] = PrependZeros(spl[i], len)
}
}
return spl.join(seperator || ' ');
}
};
Examples:
PrependZeros("1:2:3",2,":"); // "01:02:03"
PrependZeros(1,2); // "01"
PrependZeros(123,2); // "123"
PrependZeros("1 2 3",3); // "001 002 003"
PrependZeros("5-10-2012",2,"-"); //"05-10-2012"
You don't need regex for that. You can make a simple pad function yourself:
function pad(n) {
if (n < 10)
return "0" + n;
return n;
}
alert(pad(8));
alert(pad(11));
http://jsfiddle.net/DwnNG/
I now this is old, but I just saw this simple and clean solution used on the w3c and just had to share it somewhere.
var hours = currentDate.getHours();
(hours < 10 ? '0' : '') + hours;
How about:
x.replace(/^(\d)$/, "0$1");
I prefer Alex's way, but if you really want the regex way, you can try:
"2011-10-10 10:2:27".replace(/:(?=[^0](?::|$))/g, ":0");
That should do the work (it has an ternary operator in it, but also an regex^^)
stringDate.replace(/\d+/g, function(m) {
return parseInt(m, 10) < 10 ? "0" + m : m;
});
Are you ready for your one-liner do-it-all regex solution?
Pass in your entire string... and this will match single-digits and pad them with a 0.
"2011-10-17 10:3:7"
.replace(/(^|[^0-9])([0-9])(?=($|[^0-9]))/ig, function ($0, $1) {
return $0[0] + '0' + $0[1];
});
//returns "2011-10-17 10:03:07"
This would be made so much easier if JavaScript supported lookbehind-assertions.
%0{length} => %05 => 1=00001, 2=00002,... 55=00055,...
This question already has answers here:
How to format numbers as currency strings
(67 answers)
Closed 3 years ago.
I want to format numbers using JavaScript.
For example:
10 => 10.00
100 => 100.00
1000 => 1,000.00
10000 => 10,000.00
100000 => 100,000.00
If you want to use built-in code, you can use toLocaleString() with minimumFractionDigits.
Browser compatibility for the extended options on toLocaleString() was limited when I first wrote this answer, but the current status looks good. If you're using Node.js, you will need to npm install the intl package.
var value = (100000).toLocaleString(
undefined, // leave undefined to use the visitor's browser
// locale or a string like 'en-US' to override it.
{ minimumFractionDigits: 2 }
);
console.log(value);
Number formatting varies between cultures. Unless you're doing string comparison on the output,1 the polite thing to do is pick undefined and let the visitor's browser use the formatting they're most familiar with.2
// Demonstrate selected international locales
var locales = [
undefined, // Your own browser
'en-US', // United States
'de-DE', // Germany
'ru-RU', // Russia
'hi-IN', // India
'de-CH', // Switzerland
];
var n = 100000;
var opts = { minimumFractionDigits: 2 };
for (var i = 0; i < locales.length; i++) {
console.log(locales[i], n.toLocaleString(locales[i], opts));
}
If you are from a culture with a different format from those above, please edit this post and add your locale code.
1 Which you shouldn't.
2 Obviously do not use this for currency with something like {style: 'currency', currency: 'JPY'} unless you have converted to the local exchange rate. You don't want your website to tell people the price is ¥300 when it's really $300. Sometimes real e-commerce sites make this mistake.
Use
num = num.toFixed(2);
Where 2 is the number of decimal places
Edit:
Here's the function to format number as you want
function formatNumber(number)
{
number = number.toFixed(2) + '';
x = number.split('.');
x1 = x[0];
x2 = x.length > 1 ? '.' + x[1] : '';
var rgx = /(\d+)(\d{3})/;
while (rgx.test(x1)) {
x1 = x1.replace(rgx, '$1' + ',' + '$2');
}
return x1 + x2;
}
Sorce: www.mredkj.com
Short solution:
var n = 1234567890;
String(n).replace(/(.)(?=(\d{3})+$)/g,'$1,')
// "1,234,567,890"
On browsers that support the ECMAScript® 2016 Internationalization API Specification (ECMA-402), you can use an Intl.NumberFormat instance:
var nf = Intl.NumberFormat();
var x = 42000000;
console.log(nf.format(x)); // 42,000,000 in many locales
// 42.000.000 in many other locales
if (typeof Intl === "undefined" || !Intl.NumberFormat) {
console.log("This browser doesn't support Intl.NumberFormat");
} else {
var nf = Intl.NumberFormat();
var x = 42000000;
console.log(nf.format(x)); // 42,000,000 in many locales
// 42.000.000 in many other locales
}
Due to the bugs found by JasperV — good points! — I have rewritten my old code. I guess I only ever used this for positive values with two decimal places.
Depending on what you are trying to achieve, you may want rounding or not, so here are two versions split across that divide.
First up, with rounding.
I've introduced the toFixed() method as it better handles rounding to specific decimal places accurately and is well support. It does slow things down however.
This version still detaches the decimal, but using a different method than before. The w|0 part removes the decimal. For more information on that, this is a good answer. This then leaves the remaining integer, stores it in k and then subtracts it again from the original number, leaving the decimal by itself.
Also, if we're to take negative numbers into account, we need to while loop (skipping three digits) until we hit b. This has been calculated to be 1 when dealing with negative numbers to avoid putting something like -,100.00
The rest of the loop is the same as before.
function formatThousandsWithRounding(n, dp){
var w = n.toFixed(dp), k = w|0, b = n < 0 ? 1 : 0,
u = Math.abs(w-k), d = (''+u.toFixed(dp)).substr(2, dp),
s = ''+k, i = s.length, r = '';
while ( (i-=3) > b ) { r = ',' + s.substr(i, 3) + r; }
return s.substr(0, i + 3) + r + (d ? '.'+d: '');
};
In the snippet below you can edit the numbers to test yourself.
function formatThousandsWithRounding(n, dp){
var w = n.toFixed(dp), k = w|0, b = n < 0 ? 1 : 0,
u = Math.abs(w-k), d = (''+u.toFixed(dp)).substr(2, dp),
s = ''+k, i = s.length, r = '';
while ( (i-=3) > b ) { r = ',' + s.substr(i, 3) + r; }
return s.substr(0, i + 3) + r + (d ? '.'+d: '');
};
var dp;
var createInput = function(v){
var inp = jQuery('<input class="input" />').val(v);
var eql = jQuery('<span> = </span>');
var out = jQuery('<div class="output" />').css('display', 'inline-block');
var row = jQuery('<div class="row" />');
row.append(inp).append(eql).append(out);
inp.keyup(function(){
out.text(formatThousandsWithRounding(Number(inp.val()), Number(dp.val())));
});
inp.keyup();
jQuery('body').append(row);
return inp;
};
jQuery(function(){
var numbers = [
0, 99.999, -1000, -1000000, 1000000.42, -1000000.57, -1000000.999
], inputs = $();
dp = jQuery('#dp');
for ( var i=0; i<numbers.length; i++ ) {
inputs = inputs.add(createInput(numbers[i]));
}
dp.on('input change', function(){
inputs.keyup();
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input id="dp" type="range" min="0" max="5" step="1" value="2" title="number of decimal places?" />
Now the other version, without rounding.
This takes a different route and attempts to avoid mathematical calculation (as this can introduce rounding, or rounding errors). If you don't want rounding, then you are only dealing with things as a string i.e. 1000.999 converted to two decimal places will only ever be 1000.99 and not 1001.00.
This method avoids using .split() and RegExp() however, both of which are very slow in comparison. And whilst I learned something new from Michael's answer about toLocaleString, I also was surprised to learn that it is — by quite a way — the slowest method out of them all (at least in Firefox and Chrome; Mac OSX).
Using lastIndexOf() we find the possibly existent decimal point, and from there everything else is pretty much the same. Save for the padding with extra 0s where needed. This code is limited to 5 decimal places. Out of my test this was the faster method.
var formatThousandsNoRounding = function(n, dp){
var e = '', s = e+n, l = s.length, b = n < 0 ? 1 : 0,
i = s.lastIndexOf('.'), j = i == -1 ? l : i,
r = e, d = s.substr(j+1, dp);
while ( (j-=3) > b ) { r = ',' + s.substr(j, 3) + r; }
return s.substr(0, j + 3) + r +
(dp ? '.' + d + ( d.length < dp ?
('00000').substr(0, dp - d.length):e):e);
};
var formatThousandsNoRounding = function(n, dp){
var e = '', s = e+n, l = s.length, b = n < 0 ? 1 : 0,
i = s.lastIndexOf('.'), j = i == -1 ? l : i,
r = e, d = s.substr(j+1, dp);
while ( (j-=3) > b ) { r = ',' + s.substr(j, 3) + r; }
return s.substr(0, j + 3) + r +
(dp ? '.' + d + ( d.length < dp ?
('00000').substr(0, dp - d.length):e):e);
};
var dp;
var createInput = function(v){
var inp = jQuery('<input class="input" />').val(v);
var eql = jQuery('<span> = </span>');
var out = jQuery('<div class="output" />').css('display', 'inline-block');
var row = jQuery('<div class="row" />');
row.append(inp).append(eql).append(out);
inp.keyup(function(){
out.text(formatThousandsNoRounding(Number(inp.val()), Number(dp.val())));
});
inp.keyup();
jQuery('body').append(row);
return inp;
};
jQuery(function(){
var numbers = [
0, 99.999, -1000, -1000000, 1000000.42, -1000000.57, -1000000.999
], inputs = $();
dp = jQuery('#dp');
for ( var i=0; i<numbers.length; i++ ) {
inputs = inputs.add(createInput(numbers[i]));
}
dp.on('input change', function(){
inputs.keyup();
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input id="dp" type="range" min="0" max="5" step="1" value="2" title="number of decimal places?" />
I'll update with an in-page snippet demo shortly, but for now here is a fiddle:
https://jsfiddle.net/bv2ort0a/2/
Old Method
Why use RegExp for this? — don't use a hammer when a toothpick will do i.e. use string manipulation:
var formatThousands = function(n, dp){
var s = ''+(Math.floor(n)), d = n % 1, i = s.length, r = '';
while ( (i -= 3) > 0 ) { r = ',' + s.substr(i, 3) + r; }
return s.substr(0, i + 3) + r +
(d ? '.' + Math.round(d * Math.pow(10, dp || 2)) : '');
};
walk through
formatThousands( 1000000.42 );
First strip off decimal:
s = '1000000', d = ~ 0.42
Work backwards from the end of the string:
',' + '000'
',' + '000' + ',000'
Finalise by adding the leftover prefix and the decimal suffix (with rounding to dp no. decimal points):
'1' + ',000,000' + '.42'
fiddlesticks
http://jsfiddle.net/XC3sS/
Use the Number function toFixed and this function to add the commas.
function addCommas(nStr)
{
nStr += '';
var x = nStr.split('.');
var x1 = x[0];
var x2 = x.length > 1 ? '.' + x[1] : '';
var rgx = /(\d+)(\d{3})/;
while (rgx.test(x1)) {
x1 = x1.replace(rgx, '$1' + ',' + '$2');
}
return x1 + x2;
}
n = 10000;
r = n.toFixed(2); //10000.00
addCommas(r); // 10,000.00
http://www.mredkj.com/javascript/numberFormat.html
I think with this jQuery-numberformatter you could solve your problem.
Of course, this is assuming that you don't have problem with using jQuery in your project. Please notice that the functionality is tied to the blur event.
$("#salary").blur(function(){
$(this).parseNumber({format:"#,###.00", locale:"us"});
$(this).formatNumber({format:"#,###.00", locale:"us"});
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script src="https://cdn.jsdelivr.net/gh/timdown/jshashtable/hashtable.js"></script>
<script src="https://cdn.jsdelivr.net/gh/hardhub/jquery-numberformatter/src/jquery.numberformatter.js"></script>
<input type="text" id="salary">
You may want to consider using toLocaleString()
Working Example:
const number = 1234567890.123;
console.log(number.toLocaleString('en-US')); // US format
console.log(number.toLocaleString('en-IN')); // Indian format
Tested in Chrome v60 and v88
Source: Number.prototype.toLocaleString() | MDN
function numberWithCommas(x) {
x=String(x).toString();
var afterPoint = '';
if(x.indexOf('.') > 0)
afterPoint = x.substring(x.indexOf('.'),x.length);
x = Math.floor(x);
x=x.toString();
var lastThree = x.substring(x.length-3);
var otherNumbers = x.substring(0,x.length-3);
if(otherNumbers != '')
lastThree = ',' + lastThree;
return otherNumbers.replace(/\B(?=(\d{2})+(?!\d))/g, ",") + lastThree + afterPoint;
}
console.log(numberWithCommas(100000));
console.log(numberWithCommas(10000000));
Output
1,00,000
1,00,00,000
This is an article about your problem. Adding a thousands-seperator is not built in to JavaScript, so you'll have to write your own function like this (example taken from the linked page):
function addSeperator(nStr){
nStr += '';
x = nStr.split('.');
x1 = x[0];
x2 = x.length > 1 ? '.' + x[1] : '';
var rgx = /(\d+)(\d{3})/;
while (rgx.test(x1)) {
x1 = x1.replace(rgx, '$1' + ',' + '$2');
}
return x1 + x2;
}
Or you could use the sugar.js library, and the format method:
format( place = 0 , thousands = ',' , decimal = '.' ) Formats the number to a readable string. If place is undefined, will automatically
determine the place. thousands is the character used for the thousands
separator. decimal is the character used for the decimal point.
Examples:
(56782).format() > "56,782"
(56782).format(2) > "56,782.00"
(4388.43).format(2, ' ') > "4 388.43"
(4388.43).format(3, '.', ',') > "4.388,430"
Let me also throw my solution in here. I've commented each line for ease of reading and also provided some examples, so it may look big.
function format(number) {
var decimalSeparator = ".";
var thousandSeparator = ",";
// make sure we have a string
var result = String(number);
// split the number in the integer and decimals, if any
var parts = result.split(decimalSeparator);
// if we don't have decimals, add .00
if (!parts[1]) {
parts[1] = "00";
}
// reverse the string (1719 becomes 9171)
result = parts[0].split("").reverse().join("");
// add thousand separator each 3 characters, except at the end of the string
result = result.replace(/(\d{3}(?!$))/g, "$1" + thousandSeparator);
// reverse back the integer and replace the original integer
parts[0] = result.split("").reverse().join("");
// recombine integer with decimals
return parts.join(decimalSeparator);
}
document.write("10 => " + format(10) + "<br/>");
document.write("100 => " + format(100) + "<br/>");
document.write("1000 => " + format(1000) + "<br/>");
document.write("10000 => " + format(10000) + "<br/>");
document.write("100000 => " + format(100000) + "<br/>");
document.write("100000.22 => " + format(100000.22) + "<br/>");
This will get you your comma seperated values as well as add the fixed notation to the end.
nStr="1000";
nStr += '';
x = nStr.split('.');
x1 = x[0];
x2 = x.length > 1 ? '.' + x[1] : '';
var rgx = /(\d+)(\d{3})/;
while (rgx.test(x1)) {
x1 = x1.replace(rgx, '$1' + ',' + '$2');
}
commaSeperated = x1 + x2 + ".00";
alert(commaSeperated);
Source
If you're using jQuery, you could use the format or number format plugins.
function formatNumber1(number) {
var comma = ',',
string = Math.max(0, number).toFixed(0),
length = string.length,
end = /^\d{4,}$/.test(string) ? length % 3 : 0;
return (end ? string.slice(0, end) + comma : '') + string.slice(end).replace(/(\d{3})(?=\d)/g, '$1' + comma);
}
function formatNumber2(number) {
return Math.max(0, number).toFixed(0).replace(/(?=(?:\d{3})+$)(?!^)/g, ',');
}
Source: http://jsperf.com/number-format
This is about 3 times faster version of the accepted answer. It doesn't create array and avoids object creation and string concatenation for whole numbers at the end. This might be useful if you render lots of values e.g. in a table.
function addThousandSeparators(number) {
var whole, fraction
var decIndex = number.lastIndexOf('.')
if (decIndex > 0) {
whole = number.substr(0, decIndex)
fraction = number.substr(decIndex)
} else {
whole = number
}
var rgx = /(\d+)(\d{3})/
while (rgx.test(whole)) {
whole = whole.replace(rgx, '$1' + ',' + '$2')
}
return fraction ? whole + fraction : whole
}
function formatThousands(n,dp,f) {
// dp - decimal places
// f - format >> 'us', 'eu'
if (n == 0) {
if(f == 'eu') {
return "0," + "0".repeat(dp);
}
return "0." + "0".repeat(dp);
}
/* round to 2 decimal places */
//n = Math.round( n * 100 ) / 100;
var s = ''+(Math.floor(n)), d = n % 1, i = s.length, r = '';
while ( (i -= 3) > 0 ) { r = ',' + s.substr(i, 3) + r; }
var a = s.substr(0, i + 3) + r + (d ? '.' + Math.round((d+1) * Math.pow(10,dp)).toString().substr(1,dp) : '');
/* change format from 20,000.00 to 20.000,00 */
if (f == 'eu') {
var b = a.toString().replace(".", "#");
b = b.replace(",", ".");
return b.replace("#", ",");
}
return a;
}