Generate a UUID on iOS from Swift

312
votes

In my iOS Swift app I want to generate random UUID (GUID) strings for use as a table key, and this snippet appears to work:

let uuid = CFUUIDCreateString(nil, CFUUIDCreate(nil))

Is this safe?

Or is there perhaps a better (recommended) approach?

6
iosswiftcocoa-touchguidcore-foundation
You can also use let uuid = NSUUID.UUID().UUIDString – Yatheesha
Take a look at stackoverflow.com/questions/3773776/… – Yatheesha
Instead of your edit, please accept one of the answers. Since you're doing it the way @AhemdAlHafoudh proposed, I suggest you accept his answer. – DarkDust
Ah, got it. Thanks... done. – zacjordaan

6 Answers

658
votes

Try this one:

let uuid = NSUUID().uuidString
print(uuid)

Swift 3/4/5

let uuid = UUID().uuidString
print(uuid)
29
votes

You could also just use the NSUUID API:

let uuid = NSUUID()

If you want to get the string value back out, you can use uuid.UUIDString.

Note that NSUUID is available from iOS 6 and up.

20
votes

For Swift 4;

let uuid = NSUUID().uuidString.lowercased()
14
votes

For Swift 3, many Foundation types have dropped the 'NS' prefix, so you'd access it by UUID().uuidString.

9
votes

Also you can use it lowercase under below

let uuid = NSUUID().UUIDString.lowercaseString
print(uuid)

Output

68b696d7-320b-4402-a412-d9cee10fc6a3

Thank you !

7
votes

Each time the same will be generated:

if let uuid = UIDevice.current.identifierForVendor?.uuidString {
    print(uuid)
}

Each time a new one will be generated:

let uuid = UUID().uuidString
print(uuid)