run<T> static method

Future<T> run<T>(Future<T> action(), { int attempts = 3, Duration delay = const Duration(seconds: 1), double factor = 2, Duration maxDelay = const Duration(seconds: 30), bool retryIf(Object error)?, void onRetry(Object error, int attempt)?, })

Calls action until it succeeds or attempts run out, then rethrows the last error; retryIf can stop early.

Implementation

static Future<T> run<T>(
  Future<T> Function() action, {
  int attempts = 3,
  Duration delay = const Duration(seconds: 1),
  double factor = 2,
  Duration maxDelay = const Duration(seconds: 30),
  bool Function(Object error)? retryIf,
  void Function(Object error, int attempt)? onRetry,
}) async {
  Duration wait = delay;
  for (int attempt = 1; ; attempt++) {
    try {
      return await action();
    } catch (error) {
      if (attempt >= attempts || (retryIf != null && !retryIf(error))) rethrow;
      onRetry?.call(error, attempt);
      await Future<void>.delayed(wait);
      final int next = (wait.inMilliseconds * factor).round();
      wait = Duration(milliseconds: min(next, maxDelay.inMilliseconds));
    }
  }
}