How to catch exception in flutter?

Question

This is my exception class. Exception class has been implemented by the abstract exception class of flutter. Am I missing something?

class FetchDataException implements Exception {
 final _message;
 FetchDataException([this._message]);

String toString() {
if (_message == null) return "Exception";
  return "Exception: $_message";
 }
}


void loginUser(String email, String password) {
  _data
    .userLogin(email, password)
    .then((user) => _view.onLoginComplete(user))
    .catchError((onError) => {
       print('error caught');
       _view.onLoginError();
    });
}

Future < User > userLogin(email, password) async {
  Map body = {
    'username': email,
    'password': password
  };
  http.Response response = await http.post(apiUrl, body: body);
  final responseBody = json.decode(response.body);
  final statusCode = response.statusCode;
  if (statusCode != HTTP_200_OK || responseBody == null) {
    throw new FetchDataException(
      "An error occured : [Status Code : $statusCode]");
   }
  return new User.fromMap(responseBody);
}

CatchError doesn't catch the error when the status is not 200. In short error caught is not printed.

Your question `How to catch exception in flutter?` is general, but your code is personal. Can someone explain how to catch and print errors in Flutter??? — , Jul 06 '22 at 19:25

Günter Zöchbauer · Accepted Answer · 2018-10-09T17:22:22.813

74

Try

void loginUser(String email, String password) async {
  try {
    var user = await _data
      .userLogin(email, password);
    _view.onLoginComplete(user);
      });
  } on FetchDataException catch(e) {
    print('error caught: $e');
    _view.onLoginError();
  }
}

catchError is sometimes a bit tricky to get right. With async/await you can use try/catch like with sync code and it is usually much easier to get right.

edited Oct 09 '18 at 17:22

answered Oct 09 '18 at 13:01

Günter Zöchbauer

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1

Günter, I think we can't use `await` after method parameters ends. You should use `async` there. I am not having IDE right now, so can't check if `await` is valid there. – CopsOnRoad Oct 09 '18 at 17:21
1

Thanks, should have been `async` – Günter Zöchbauer Oct 09 '18 at 17:22

score 18 · Answer 2 · answered May 31 '21 at 14:45

Let's say this is your function which throws an exception:

Future<void> foo() async {
  throw Exception('FooException');
}

You can either use try-catch block or catchError on the Future since both do the same thing.

Using try-catch

try {
  await foo();
} on Exception catch (e) {
  print(e); // Only catches an exception of type `Exception`.
} catch (e) {
  print(e); // Catches all types of `Exception` and `Error`.
}

Use catchError
```
await foo().catchError(print);
```

score 2 · Answer 3 · answered Oct 30 '21 at 03:08

I was trying to find this answer when got to this page, hope it helps: https://stackoverflow.com/a/57736915/12647239

Basicly i was just trying to catch an error message from a method, but i was calling

throw Exception("message")

And in "catchError" i was getting "Exception: message" instead of "message".

catchError(
  (error) => print(error)
);

fixed with the return in the above reference

score 1 · Answer 4 · answered Aug 27 '20 at 10:37

Future < User > userLogin(email, password) async { try {
  Map body = {
    'username': email,
    'password': password
  };
  http.Response response = await http.post(apiUrl, body: body);
  final responseBody = json.decode(response.body);
  final statusCode = response.statusCode;
  if (statusCode != HTTP_200_OK || responseBody == null) {
    throw new FetchDataException(
      "An error occured : [Status Code : $statusCode]");
   }
  return new User.fromMap(responseBody); }
   catch (e){
    print(e.toString());
}

score 0 · Answer 5 · answered Aug 17 '23 at 04:44

You can throw exception as below as well.

 Future.error('Location permissions are denied');

More important, Sometimes you may face to a situation where you want to take action on certain exceptions.

Example:

  _getCurrentLocation() async {
    try {
      LocationPermission permission = await Geolocator.checkPermission();
      if (permission == LocationPermission.denied) {
        permission = await Geolocator.requestPermission();
        if (permission == LocationPermission.denied) {
          throw Exception("Location permissions are denied");
          //Future.error('Location permissions are denied');
        }
      }

      if (permission == LocationPermission.deniedForever) {
        throw Exception("Location permissions permanently denied.");
        //Future.error('Location permissions permanently denied.');
      }

      Position pos = await Geolocator.getCurrentPosition();
      _currentPosition = LatLng(pos.latitude, pos.longitude);
    } on Exception catch (e) {
      String message = e.toString();
      if (message == "Location permissions are denied") {
        print("Action on denied location");
        return;
      }
      if (message == "Location permissions permanently denied.") {
        print("Action on permanently denied location");
        return;
      }

    }

How to catch exception in flutter?

5 Answers5

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