Flutter Riverpod 消除多个依赖提供程序

Question

正如 Riverpod 文档中所述，Riverpod 提供程序可以监视其他提供程序来创建“处理管道”。

我有这样的事情：

final prov = Provider<String>((ref){

    final w1 = ref.watch(prov1);
    final w2 = ref.watch(prov2);
    final w3 = ref.watch(prov3);
    final w4 = ref.watch(prov4);

    final complex = expensiveFunction(w1,w2,w3,w4);

    return complex;
});

prov1可以prov4通过 UI 的各个部分单独修改，但某些 UI 操作会导致其中部分或全部快速连续更改。

我怎样才能消除呼叫expensiveFunction()直到w1..w4在 2 秒内都没有改变？

Answer 1

根据Riverpod 包作者的推文，您可以执行以下操作：

/// An extension on [Ref] with helpful methods to add a debounce.
extension RefDebounceExtension on Ref {
  /// Delays an execution by a bit such that if a dependency changes multiple
  /// time rapidly, the rest of the code is only run once.
  Future<void> debounce(Duration duration) {
    final completer = Completer<void>();
    final timer = Timer(duration, () {
      if (!completer.isCompleted) completer.complete();
    });
    onDispose(() {
      timer.cancel();
      if (!completer.isCompleted) {
        completer.completeError(StateError('Cancelled'));
      }
    });
    return completer.future;
  }
}

final prov = FutureProvider<String>((ref) async {

    final w1 = ref.watch(prov1);
    final w2 = ref.watch(prov2);
    final w3 = ref.watch(prov3);
    final w4 = ref.watch(prov4);

    await debounce(Duration(seconds: 2));

    final complex = expensiveFunction(w1,w2,w3,w4);

    return complex;
});