how can i generate randome number using replace function javascript? [duplicate]

Question

i'm new to javascript i need to generate random number using replace it is possible i try

function generateUUID() {
    var dt = new Date().getTime()
    var uuid = 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace('/[xy]/g', function (c) {
        var r = (dt + Math.random() * 16) % 16 | 0
        dt = Math.floor(dt / 16)
        return (c == 'x' ? r : (r & 0x3 | 0x8)).toString(16)
    })
    return uuid
}

this function always return xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx What am i missing ?

does the string '/[xy]/g' exist? no ... you want a regular expression ... /[xy]/g — Jaromanda X
– Jaromanda X, Commented Jun 30, 2021 at 10:10
use regular expression format properly /[xy]/g instead of '/[xy]/g' — subrahmanya bhat
– subrahmanya bhat, Commented Jun 30, 2021 at 10:14

Terry Lennox · Accepted Answer · 2021-06-30 10:23:09Z

1

You're using a string rather than a RegExp in the replace function, this should be a simple fix, e.g. just change

.replace('/[xy]/g',

to

.replace(/[xy]/g,

function generateUUID() {
    var dt = new Date().getTime()
    var uuid = 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function (c) {
        var r = (dt + Math.random() * 16) % 16 | 0
        dt = Math.floor(dt / 16)
        return (c == 'x' ? r : (r & 0x3 | 0x8)).toString(16)
    })
    console.log("uuid",uuid)
    return uuid
}

console.log(generateUUID())

I'd also suggest looking at third-party libraries for this purpose, uuid for example, it uses cryptographically-strong random values.

function generateUUID() {
    return uuid.v4();
}
console.log(generateUUID());

<script src="https://cdnjs.cloudflare.com/ajax/libs/uuid/8.3.2/uuid.min.js" integrity="sha512-UNM1njAgOFUa74Z0bADwAq8gbTcqZC8Ej4xPSzpnh0l6KMevwvkBvbldF9uR++qKeJ+MOZHRjV1HZjoRvjDfNQ==" crossorigin="anonymous" referrerpolicy="no-referrer"></script>

edited Jun 30, 2021 at 10:23

answered Jun 30, 2021 at 10:10

Terry Lennox

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2 Comments

Brock Over a year ago

thanks @Terry Lennox

Terry Lennox Over a year ago

No problem, glad to be of some help!

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