'How to generate unique id in Dart

I write websocket chat. How to generate unique id for user?

now i use this code:

id = new DateTime.now().millisecondsSinceEpoch;

is there any more neat solution?

Solution 1:^[1]

In year 2020 you can do UniqueKey(); which is a built in class:

https://api.flutter.dev/flutter/widgets/UniqueKey-class.html

Note

A key that is only equal to itself.

This cannot be created with a const constructor because that implies that all instantiated keys would be the same instance and therefore not be unique.

Solution 2:^[2]

Besides from uuid, you can also try this to generate small unique keys:

https://pub.dev/packages/nanoid

They even have a collision calculator:

https://zelark.github.io/nano-id-cc/

Solution 3:^[3]

I use microseconds instead of milliseconds, which is much more accurate and there is no need to add any package.

 String idGenerator() {
    final now = DateTime.now();
    return now.microsecondsSinceEpoch.toString();
  }

Solution 4:^[4]

If you like MongoDB style ids you could consider this small package that will help create the object id:

https://pub.dev/packages/crossplat_objectid

import 'package:bson_objectid/bson_objectid.dart';

main() {
  ObjectId id1 = new ObjectId();
  print(id1.toHexString());

  ObjectId id2 = new ObjectId.fromHexString('54495ad94c934721ede76d90');
  print(id2.timestamp);
  print(id2.machineId);
  print(id2.processId);
  print(id2.counter);
}

Solution 5:^[5]

There is also https://pub.dev/packages/xid which is lock free and has a Unicity guaranteed for 16,777,216 (24 bits) unique ids per second and per host/process

import 'package:xid/xid.dart';

void main() {
  var xid = Xid();
  print('generated id: $xid');

}

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution	Source
Solution 1	BIS Tech
Solution 2	theredforest
Solution 3	dev001
Solution 4	34m0
Solution 5	Bwire

'How to generate unique id in Dart

Solution 1:[1]

Solution 2:[2]

Solution 3:[3]

Solution 4:[4]

Solution 5:[5]