如何从Firebase获取对象或异常？

Question

我有以下通用类：

public class DataOrException<T, E extends Exception> {
    public T data;
    public E exception;
}

而且我正在从Firebase身份验证，FirebaseUser对象或异常中获取：

MutableLiveData<DataOrException<FirebaseUser, Exception>> signIn(String email, String password) {
    MutableLiveData<DataOrException<FirebaseUser, Exception>> liveData = new MutableLiveData<>();
    auth.signInWithEmailAndPassword(email, password).addOnCompleteListener(task -> {
        if (task.isSuccessful()) {
            FirebaseUser firebaseUser = auth.getCurrentUser();
            liveData.setValue(firebaseUser);
        } else {
            liveData.setValue(authTask.getException());
        }
    });
    return liveData;
}

但出现以下错误：

setValue(com.example.DataOrException<com.google.firebase.auth.FirebaseUser, java.lang.Exception>)
in MutableLiveData cannot be applied to (java.lang.Exception)

我应该传递给setValue（）以便它不再抱怨吗？

Answer 1

您的MutableLiveData需要一个DataOrException<FirebaseUser, Exception>类型的值，因此您需要构造一个并传递它。

DataOrException<FirebaseUser, Exception> value = new DataOrException<>()
if (task.isSuccessful()) {
    value.data = auth.getCurrentUser();
} else {
    value.exception = authTask.getException();
}
liveData.setValue(value);