Create a random token in Javascript based on user details

Question

I want to create a random string (token) which can be used to identify a user whilst avoiding any potential conflicts with any other users\' tokens.

What I was thinking

Answer 1

This function allows you to set the token length and allowed characters.

function generate_token(length){
    //edit the token allowed characters
    var a = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890".split("");
    var b = [];  
    for (var i=0; i<length; i++) {
        var j = (Math.random() * (a.length-1)).toFixed(0);
        b[i] = a[j];
    }
    return b.join("");
}

Simply call generate_token

generate_token(32); //returns "qweQj4giRJSdMNzB8g1XIa6t3YtRIHPH"

Answer 2

Checkout crypto.js project. Its a collection of cryptographic algorithms. The project has separate js files for each hashing algorithms.

Answer 3

//length: defines the length of characters to express in the string

const rand=()=>Math.random(0).toString(36).substr(2);
const token=(length)=>(rand()+rand()+rand()+rand()).substr(0,length);

console.log(token(40));
//example1:  token(10) => result: tsywlmdqu6
//example2:  token(40) => result: m4vni14mtln2547gy54ksclhcv0dj6tp9fhs1k10

Answer 4

You could generate a random number and convert it to base 36 (0-9a-z):

var rand = function() {
    return Math.random().toString(36).substr(2); // remove `0.`
};

var token = function() {
    return rand() + rand(); // to make it longer
};

token(); // "bnh5yzdirjinqaorq0ox1tf383nb3xr"

Answer 5

I use an approach similar to Kareem's, but with fewer function calls and built-in array operations for a big boost in performance.

According to a performance test, this method also outperforms the accepted answer by a small margin. Moreover it provides a parameter n to generate any size token length from a white list of acceptable characters. It's flexible and performs well.

function generateToken(n) {
    var chars = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789';
    var token = '';
    for(var i = 0; i < n; i++) {
        token += chars[Math.floor(Math.random() * chars.length)];
    }
    return token;
}

Answer 6

It is very unlikely, but Math.random() could return 0.0. In that case, the solution of pimvdb would return "" (empty string). So, here is another solution, which returns in every case a random base36 with 10 chars length:

function generateToken() {
    Math.floor(1000000000000000 + Math.random() * 9000000000000000)
          .toString(36).substr(0, 10)
}