I want a 5 character string composed of characters picked randomly from the set [azAZ09]
.
What's the best way to do this with JavaScript?
I think this will work for you:
function makeid(length) {
var result = '';
var characters = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
var charactersLength = characters.length;
for ( var i = 0; i < length; i++ ) {
result += characters.charAt(Math.floor(Math.random() * charactersLength));
}
return result;
}
console.log(makeid(5));
let r = Math.random().toString(36).substring(7);
console.log("random", r);
Note: The above algorithm has the following weaknesses:
Math.random()
may produce predictable ("randomlooking" but not really random) output depending on the implementation. The resulting string is not suitable when you need to guarantee uniqueness or unpredictability.Math.random is bad for this kind of thing
Option 1
If you're able to do this serverside, just use the crypto module 
var crypto = require("crypto");
var id = crypto.randomBytes(20).toString('hex');
// "bb5dc8842ca31d4603d6aa11448d1654"
The resulting string will be twice as long as the random bytes you generate; each byte encoded to hex is 2 characters. 20 bytes will be 40 characters of hex.
Option 2
If you have to do this clientside, perhaps try the uuid module 
var uuid = require("uuid");
var id = uuid.v4();
// "110ec58aa0f24ac48393c866d813b8d1"
Option 3
If you have to do this clientside and you don't have to support old browsers, you can do it without dependencies 
// dec2hex :: Integer > String
// i.e. 0255 > '00''ff'
function dec2hex (dec) {
return ('0' + dec.toString(16)).substr(2)
}
// generateId :: Integer > String
function generateId (len) {
var arr = new Uint8Array((len  40) / 2)
window.crypto.getRandomValues(arr)
return Array.from(arr, dec2hex).join('')
}
console.log(generateId())
// "82defcf324571e70b0521d79cce2bf3fffccd69"
console.log(generateId(20))
// "c1a050a4cd1556948d41"
For more information on crypto.getRandomValues

The
crypto.getRandomValues()
method lets you get cryptographically strong random values. The array given as the parameter is filled with random numbers (random in its cryptographic meaning).
Here's a little console example 
> var arr = new Uint8Array(4) # make array of 4 bytes (values 0255)
> arr
Uint8Array(4) [ 0, 0, 0, 0 ]
> window.crypto
Crypto { subtle: SubtleCrypto }
> window.crypto.getRandomValues()
TypeError: Crypto.getRandomValues requires at least 1 argument, but only 0 were passed
> window.crypto.getRandomValues(arr)
Uint8Array(4) [ 235, 229, 94, 228 ]
For IE11 support you can use 
(window.crypto  window.msCrypto).getRandomValues(arr)
For browser coverage see https://caniuse.com/#feat=getrandomvalues
Returns exactly 5 random characters, as opposed to some of the top rated answers found here.
Math.random().toString(36).substr(2, 5);
Here's an improvement on doubletap's excellent answer. The original has two drawbacks which are addressed here:
First, as others have mentioned, it has a small probability of producing short strings or even an empty string (if the random number is 0), which may break your application. Here is a solution:
(Math.random().toString(36)+'00000000000000000').slice(2, N+2)
Second, both the original and the solution above limit the string size N to 16 characters. The following will return a string of size N for any N (but note that using N > 16 will not increase the randomness or decrease the probability of collisions):
Array(N+1).join((Math.random().toString(36)+'00000000000000000').slice(2, 18)).slice(0, N)
Explanation:
Further thoughts:
Update:
Here are a couple other functionalstyle oneliners I came up with. They differ from the solution above in that:
So, say your alphabet of choice is
var s = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
Then these two are equivalent to each other, so you can pick whichever is more intuitive to you:
Array(N).join().split(',').map(function() { return s.charAt(Math.floor(Math.random() * s.length)); }).join('');
and
Array.apply(null, Array(N)).map(function() { return s.charAt(Math.floor(Math.random() * s.length)); }).join('');
Edit:
I seems like qubyte and Martijn de Milliano came up with solutions similar to the latter (kudos!), which I somehow missed. Since they don't look as short at a glance, I'll leave it here anyway in case someone really wants a oneliner :)
Also, replaced 'new Array' with 'Array' in all solutions to shave off a few more bytes.
The most compact solution, because slice
is shorter than substring
. Subtracting from the end of the string allows to avoid floating point symbol generated by the random
function:
Math.random().toString(36).slice(5);
or even
(+new Date).toString(36).slice(5);
Update: Added one more approach using btoa
method:
btoa(Math.random()).slice(0, 5);
btoa(+new Date).slice(7, 2);
btoa(+new Date).substr(7, 5);
// Using Math.random and Base 36:
console.log(Math.random().toString(36).slice(5));
// Using new Date and Base 36:
console.log((+new Date).toString(36).slice(5));
// Using Math.random and Base 64 (btoa):
console.log(btoa(Math.random()).slice(0, 5));
// Using new Date and Base 64 (btoa):
console.log(btoa(+new Date).slice(7, 2));
console.log(btoa(+new Date).substr(7, 5));
Something like this should work
function randomString(len, charSet) {
charSet = charSet  'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
var randomString = '';
for (var i = 0; i < len; i++) {
var randomPoz = Math.floor(Math.random() * charSet.length);
randomString += charSet.substring(randomPoz,randomPoz+1);
}
return randomString;
}
Call with default charset [azAZ09] or send in your own:
var randomValue = randomString(5);
var randomValue = randomString(5, 'PICKCHARSFROMTHISSET');
function randomstring(L) {
var s = '';
var randomchar = function() {
var n = Math.floor(Math.random() * 62);
if (n < 10) return n; //110
if (n < 36) return String.fromCharCode(n + 55); //AZ
return String.fromCharCode(n + 61); //az
}
while (s.length < L) s += randomchar();
return s;
}
console.log(randomstring(5));
A newer version with es6 spread operator:
[...Array(30)].map(() => Math.random().toString(36)[2]).join('')
30
is arbitrary number, you can pick any token length you want36
is the maximum radix number you can pass to numeric.toString(), which means all numbers and az lowercase letters2
is used to pick the 3th index from the random string which looks like this: "0.mfbiohx64i"
, we could take any index after 0.
/**
* Pseudorandom string generator
* http://stackoverflow.com/a/27872144/383904
* Default: return a random alphanumeric string
*
* @param {Integer} len Desired length
* @param {String} an Optional (alphanumeric), "a" (alpha), "n" (numeric)
* @return {String}
*/
function randomString(len, an) {
an = an && an.toLowerCase();
var str = "",
i = 0,
min = an == "a" ? 10 : 0,
max = an == "n" ? 10 : 62;
for (; i++ < len;) {
var r = Math.random() * (max  min) + min << 0;
str += String.fromCharCode(r += r > 9 ? r < 36 ? 55 : 61 : 48);
}
return str;
}
console.log(randomString(10)); // i.e: "4Z8iNQag9v"
console.log(randomString(10, "a")); // i.e: "aUkZuHNcWw"
console.log(randomString(10, "n")); // i.e: "9055739230"
While the above uses additional checks for the desired A/N, A, N output, let's break it down the to the essentials (AlphaNumeric only) for a better understanding:
var str = "";
to concatenate random charactersrand
index number from 0 to 61 (0..9+A..Z+a..z = 62)rand
(since it's 0..61) incrementing it by some number (see examples below) to get back the right CharCode
number and the related Character.str
a String.fromCharCode( incremented rand )
Let's picture the ASCII Character table ranges:
_____0....9______A..........Z______a..........z___________ Character
 10   26   26  Tot = 62 characters
48....57 65..........90 97..........122 CharCode ranges
Math.floor( Math.random * 62 )
gives a range from 0..61
(what we need).
Let's fix the random to get the correct charCode ranges:
 rand  charCode  (0..61)rand += fix = charCode ranges 
+++++
0..9  0..9  48..57  rand += 48 = 48..57 
A..Z  10..35  65..90  rand += 55 /* 9035 = 55 */ = 65..90 
a..z  36..61  97..122  rand += 61 /* 12261 = 61 */ = 97..122 
The conditional operation logic from the table above:
rand += rand>9 ? ( rand<36 ? 55 : 61 ) : 48 ;
// rand += true ? ( true ? 55 else 61 ) else 48 ;
From the explanation above, here's the resulting alphanumeric snippet:
function randomString(len) {
var str = ""; // String result
for (var i = 0; i < len; i++) { // Loop `len` times
var rand = Math.floor(Math.random() * 62); // random: 0..61
var charCode = rand += rand > 9 ? (rand < 36 ? 55 : 61) : 48; // Get correct charCode
str += String.fromCharCode(charCode); // add Character to str
}
return str; // After all loops are done, return the concatenated string
}
console.log(randomString(10)); // i.e: "7GL9F0ne6t"
Or if you will:
const randomString = (n, r='') => {
while (n) r += String.fromCharCode((r=Math.random()*620, r+=r>9?(r<36?55:61):48));
return r;
};
console.log(randomString(10))
The simplest way is:
(new Date%9e6).toString(36)
This generate random strings of 5 characters based on the current time. Example output is 4mtxj
or 4mv90
or 4mwp1
The problem with this is that if you call it two times on the same second, it will generate the same string.
The safer way is:
(0Math.random()*9e6).toString(36)
This will generate a random string of 4 or 5 characters, always diferent. Example output is like 30jzm
or 1r591
or 4su1a
In both ways the first part generate a random number. The .toString(36)
part cast the number to a base36 (alphadecimal) representation of it.
Here are some easy one liners. Change new Array(5)
to set the length.
09az
new Array(5).join().replace(/(.$)/g, function(){return ((Math.random()*36)0).toString(36);})
09azAZ
new Array(5).join().replace(/(.$)/g, function(){return ((Math.random()*36)0).toString(36)[Math.random()<.5?"toString":"toUpperCase"]();});
I know everyone has got it right already, but i felt like having a go at this one in the most lightweight way possible(light on code, not CPU):
function rand(length, current) {
current = current ? current : '';
return length ? rand(length, "0123456789ABCDEFGHIJKLMNOPQRSTUVWXTZabcdefghiklmnopqrstuvwxyz".charAt(Math.floor(Math.random() * 60)) + current) : current;
}
console.log(rand(5));
It takes a bit of time to wrap your head around, but I think it really shows how awesome javascript's syntax is.
If you are using Lodash or Underscore, then it so simple:
var randomVal = _.sample('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789', 5).join('');
In case anyone is interested in a oneliner (although not formatted as such for your convenience) that allocates the memory at once (but note that for small strings it really does not matter) here is how to do it:
Array.apply(0, Array(5)).map(function() {
return (function(charset){
return charset.charAt(Math.floor(Math.random() * charset.length))
}('ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789'));
}).join('')
You can replace 5
by the length of the string you want. Thanks to @AriyaHidayat in this post for the solution to the map
function not working on the sparse array created by Array(5)
.
Here's the method I created.
It will create a string containing both uppercase and lowercase characters.
In addition I've included the function that will created an alphanumeric string too.
Working examples:
http://jsfiddle.net/greatbigmassive/vhsxs/ (alpha only)
http://jsfiddle.net/greatbigmassive/PJwg8/ (alphanumeric)
function randString(x){
var s = "";
while(s.length<x&&x>0){
var r = Math.random();
s+= String.fromCharCode(Math.floor(r*26) + (r>0.5?97:65));
}
return s;
}
Upgrade July 2015
This does the same thing but makes more sense and includes all letters.
var s = "";
while(s.length<x&&x>0){
v = Math.random()<0.5?32:0;
s += String.fromCharCode(Math.round(Math.random()*((122v)(97v))+(97v)));
}
To meet requirement [azAZ09] and length=5 use
btoa(Math.random()).substr(5, 5);
Lowercase letters, uppercase letters, and numbers will occur.
Assuming you use underscorejs it's possible to elegantly generate random string in just two lines:
var possible = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
var random = _.sample(possible, 5).join('');
Fast and improved algorithm. Does not guarantee uniform (see comments).
function getRandomId(length) {
if (!length) {
return '';
}
const possible =
'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
let result = '';
let array;
if ('Uint8Array' in self && 'crypto' in self && length <= 65536) {
array = new Uint8Array(length);
self.crypto.getRandomValues(array);
} else {
array = new Array(length);
for (let i = 0; i < length; i++) {
array[i] = Math.floor(Math.random() * 62);
}
}
for (let i = 0; i < length; i++) {
result += possible.charAt(array[i] % 62);
}
return result;
}
You can loop through an array of items and recursively add them to a string variable, for instance if you wanted a random DNA sequence:
function randomDNA(len) {
len = len  100
var nuc = new Array("A", "T", "C", "G")
var i = 0
var n = 0
s = ''
while (i <= len  1) {
n = Math.floor(Math.random() * 4)
s += nuc[n]
i++
}
return s
}
console.log(randomDNA(5));
The problem with responses to "I need random strings" questions (in whatever language) is practically every solution uses a flawed primary specification of string length. The questions themselves rarely reveal why the random strings are needed, but I would challenge you rarely need random strings of length, say 8. What you invariably need is some number of unique strings, for example, to use as identifiers for some purpose.
There are two leading ways to get strictly unique strings: deterministically (which is not random) and store/compare (which is onerous). What do we do? We give up the ghost. We go with probabilistic uniqueness instead. That is, we accept that there is some (however small) risk that our strings won't be unique. This is where understanding collision probability and entropy are helpful.
So I'll rephrase the invariable need as needing some number of strings with a small risk of repeat. As a concrete example, let's say you want to generate a potential of 5 million IDs. You don't want to store and compare each new string, and you want them to be random, so you accept some risk of repeat. As example, let's say a risk of less than 1 in a trillion chance of repeat. So what length of string do you need? Well, that question is underspecified as it depends on the characters used. But more importantly, it's misguided. What you need is a specification of the entropy of the strings, not their length. Entropy can be directly related to the probability of a repeat in some number of strings. String length can't.
And this is where a library like EntropyString can help. To generate random IDs that have less than 1 in a trillion chance of repeat in 5 million strings using entropystring
:
import {Random, Entropy} from 'entropystring'
const random = new Random()
const bits = Entropy.bits(5e6, 1e12)
const string = random.string(bits)
"44hTNghjNHGGRHqH9"
entropystring
uses a character set with 32 characters by default. There are other predefined characters sets, and you can specify your own characters as well. For example, generating IDs with the same entropy as above but using hex characters:
import {Random, Entropy, charSet16} from './entropystring'
const random = new Random(charSet16)
const bits = Entropy.bits(5e6, 1e12)
const string = random.string(bits)
"27b33372ade513715481f"
Note the difference in string length due to the difference in total number of characters in the character set used. The risk of repeat in the specified number of potential strings is the same. The string lengths are not. And best of all, the risk of repeat and the potential number of strings is explicit. No more guessing with string length.
function randomString (strLength, charSet) {
var result = [];
strLength = strLength  5;
charSet = charSet  'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
while (strLength) {
result.push(charSet.charAt(Math.floor(Math.random() * charSet.length)));
}
return result.join('');
}
This is as clean as it will get. It is fast too, http://jsperf.com/ayrandomstring.
I did not find a clean solution for supporting both lowercase and uppercase characters.
Lowercase only support:
Math.random().toString(36).substr(2, 5)
Building on that solution to support lowercase and uppercase:
Math.random().toString(36).substr(2, 5).split('').map(c => Math.random() < 0.5 ? c.toUpperCase() : c).join('');
Change the 5
in substr(2, 5)
to adjust to the length you need.
This works for sure
<script language="javascript" type="text/javascript">
function randomString() {
var chars = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXTZabcdefghiklmnopqrstuvwxyz";
var string_length = 8;
var randomstring = '';
for (var i=0; i<string_length; i++) {
var rnum = Math.floor(Math.random() * chars.length);
randomstring += chars.substring(rnum,rnum+1);
}
document.randform.randomfield.value = randomstring;
}
</script>
How about something like this: Date.now().toString(36)
Not very random, but short and quite unique every time you call it.
Generate 10 characters long string. Length is set by parameter (default 10).
function random_string_generator(len) {
var len = len  10;
var str = '';
var i = 0;
for(i=0; i<len; i++) {
switch(Math.floor(Math.random()*3+1)) {
case 1: // digit
str += (Math.floor(Math.random()*9)).toString();
break;
case 2: // small letter
str += String.fromCharCode(Math.floor(Math.random()*26) + 97); //'a'.charCodeAt(0));
break;
case 3: // big letter
str += String.fromCharCode(Math.floor(Math.random()*26) + 65); //'A'.charCodeAt(0));
break;
default:
break;
}
}
return str;
}
You can use coderain. It's a library to generate random codes according to given pattern. Use #
as a placeholder for upper and lowercase characters as well as digits:
var cr = new CodeRain("#####");
console.log(cr.next());
There are other placeholders like A
for uppercase letters or 9
for digits.
What may be useful is that calling .next()
will always give you a unique result so you don't have to worry about duplicates.
Here is a demo application that generates a list of unique random codes.
Full disclosure: I'm the author of coderain.
How about this compact little trick?
var possible = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
var stringLength = 5;
function pickRandom() {
return possible[Math.floor(Math.random() * possible.length)];
}
var randomString = Array.apply(null, Array(stringLength)).map(pickRandom).join('');
You need the Array.apply
there to trick the empty array into being an array of undefineds.
If you're coding for ES2015, then building the array is a little simpler:
var randomString = Array.from({ length: stringLength }, pickRandom).join('');
function randStr(len) {
let s = '';
while (s.length < len) s += Math.random().toString(36).substr(2, len  s.length);
return s;
}
// usage
console.log(randStr(50));
The benefit of this function is that you can get different length random string and it ensures the length of the string.
function randStr(len) {
let s = '';
while (len) s += String.fromCodePoint(Math.floor(Math.random() * (126  33) + 33));
return s;
}
// usage
console.log(randStr(50));
function randStr(len, chars='abc123') {
let s = '';
while (len) s += chars[Math.floor(Math.random() * chars.length)];
return s;
}
// usage
console.log(randStr(50));
console.log(randStr(50, 'abc'));
console.log(randStr(50, 'aab')); // more a than b
This one combines many of the answers give.
var randNo = Math.floor(Math.random() * 100) + 2 + "" + new Date().getTime() + Math.floor(Math.random() * 100) + 2 + (Math.random().toString(36).replace(/[^azAZ]+/g, '').substr(0, 5));
console.log(randNo);
I have been using it for 1 month with great results.
const c = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'
const s = [...Array(5)].map(_ => c[~~(Math.random()*c.length)]).join('')
One liner:
Array(15).fill(null).map(() => Math.random().toString(36).substr(2)).join('')
// Outputs: 0h61cbpw96y83qtnunwme5lxk1i70a6o5r5lckfcyh1dl9fffydcfxddd69ada9tu9jvqdx864xj1ul3wtfztmh2oz2vs3mv6ej0fe58ho1cftkjcuyl2lfkmxlwua83ibotxqc4guyuvrvtf60naob26t6swzpil
Here is my approach (with TypeScript).
I've decided to write yet another response because I didn't see any simple solution using modern js and clean code.
const DEFAULT_ALPHABET = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
function getRandomCharFromAlphabet(alphabet: string): string {
return alphabet.charAt(Math.floor(Math.random() * alphabet.length));
}
function generateId(idDesiredLength: number, alphabet = DEFAULT_ALPHABET): string {
/**
* Create nlong array and map it to random chars from given alphabet.
* Then join individual chars as string
*/
return Array.from({length: idDesiredLength}).map(() => {
return getRandomCharFromAlphabet(alphabet);
}).join('');
}
generateId(5); // jNVv7
©2020 All rights reserved.