firebase - Firebase,我如何捕捉异常并告诉用户它?
问题描述
Future<Users?>createUserWithEmailAndPassword(String email, String password) async {
final credential = await _firebaseAuth.createUserWithEmailAndPassword(
email: email,
password: password
);
return _userFromFirebase(credential.user);
}
解决方案
您可以使用 a 捕获异常并以try/catch
多种方式将其显示在 UI 上。例子
Future<Users?> createUserWithEmailAndPassword(
String email,
String password,
) async {
try {
final credential = await _firebaseAuth.createUserWithEmailAndPassword(
email: email,
password: password,
);
return _userFromFirebase(credential.user);
} on FirebaseException catch (e) {
// FirebaseException
print(e.message);
} catch (e) {
// all other exceptions
print(e);
}
}