How to return value from this function??? in flutter

Question

Future<String> uploadFile(String filePath) async {
    File file = File(filePath);
    final fileName = basename(filePath);
    try {
      firebase_storage.TaskSnapshot taskSnapshot = await firebase_storage.FirebaseStorage.instance
          .ref("news/$fileName")
          .putFile(file);
      await taskSnapshot.ref.getDownloadURL().then(
            (value) {
              print("Done: $value");
              return value;
            }

      );
      
      return "123";
    } on firebase_core.FirebaseException catch (e) {
      print(e);
      return "#";
    }
  }

uploadFile returns me 123 since the value is get printed after that, How do I return value from this function before returning "123"??

return await taskSnapshot.ref.... and of course remove return "123"; — pskink
– pskink, Commented Aug 14, 2021 at 6:28

Sajad Abdollahi · Accepted Answer · 2021-08-14 07:05:47Z

1

Instead of using .then(...), You need to wait for the getDownloadUrl() task and then return it's value.

Future<String> uploadFile(String filePath) async {
  File file = File(filePath);
  final fileName = basename(filePath);
  try {
    firebase_storage.TaskSnapshot taskSnapshot = await firebase_storage
        .FirebaseStorage.instance
        .ref("news/$fileName")
        .putFile(file);
    final url = await taskSnapshot.ref.getDownloadURL();
    return url;
  } on firebase_storage.FirebaseException catch (e) {
    print(e);
    return "#";
  }
}

answered Aug 14, 2021 at 7:05

Sajad Abdollahi

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