Flutter/Dart return Future

Question

I have a function, which I want to return a Future of String or null. For simplicity, the function will return null if delay for 1 second is success, or else it will return "failed";

However; I got the

  Future<String?> fcn(String str) {
    return Future.delayed(const Duration(seconds: 1)).then((value) {
      return null;
    }).onError((error, stackTrace) {
      return "failed"; // The return type 'String' isn't a 'FutureOr<Null>', as required by the closure's context.
    });
  }

What about `return Future.value("failed");`? – Peter Koltai Oct 18 '21 at 19:42 — Peter Koltai, Oct 18 '21 at 19:42
Use `async`/`await`. It is much more straightforward. – jamesdlin Oct 18 '21 at 20:55 — jamesdlin, Oct 18 '21 at 20:55

Sergio · Answer 1 · 2021-10-18T19:56:28.087

0

Try to use onError as parameter of then method:

Future<String?> fcn(String str) {
  return Future.delayed(const Duration(seconds: 1)).then((value) {
    return null;
  }, onError: (error, stackTrace) {
    return "failed";
  });
}

Or method catchError:

Future<String?> fcn(String str) {
  return Future.delayed(const Duration(seconds: 1)).then((value) {
    return null;
  }).catchError((error, stackTrace) {
    return "failed";
  });
}

edited Oct 18 '21 at 19:56

answered Oct 18 '21 at 19:46

Sergio

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Thank you that worked, but why is that? – Steven Z. Oct 18 '21 at 20:36
It is strange that you don't get another error: "The method 'onError' isn't defined for the type 'Future" bcz there is no `onError` method defined in `Future` class https://api.dart.dev/stable/2.14.4/dart-async/Future-class.html – Sergio Oct 18 '21 at 20:57

Flutter/Dart return Future

1 Answers1