LiveData观察失败
LiveData观察失败
提问于 2019-03-05 05:27:58
我正在尝试用MVVM启动应用程序(我是新手)，但我遇到了一些问题。
我需要获取livedata值，创建用户对象，然后在postRequest中发送该对象。
这是我的loginViewModel：
public class LoginViewModel extends AndroidViewModel {
private UserRepository userRepository;
private MutableLiveData<String> user;
private MutableLiveData<String> password;
private MutableLiveData<String> email;
private MutableLiveData<Resource<User>> userObservable;
@Inject
public LoginViewModel(@NonNull Application application){
super(application);
this.userRepository = new UserRepository();
}
public MutableLiveData<String> getUser() {
if(user == null)
user = new MutableLiveData<>();
return user;
}
public MutableLiveData<String> getPassword() {
if(password == null)
password = new MutableLiveData<>();
return password;
}
public MutableLiveData<String> getEmail() {
if(email == null)
email = new MutableLiveData<>();
return email;
}
public void loginOnClickListener(){
User user = new User(getUser().getValue(), getPassword().getValue(), getEmail().getValue());
userRepository.saveUser(user, userObservable);
}
}在我的代码片段中，每个引用值都处于观察模式：
@Override
public void onActivityCreated(Bundle savedInstanceState) {
super.onActivityCreated(savedInstanceState);
loginViewModel = ViewModelProviders.of(getActivity(), viewModelFactory).get(LoginViewModel.class);
binding.setLoginViewModel(loginViewModel);
observeViewModel();
}
private void observeViewModel(){
loginViewModel.getUser().observe(this, user -> {
String coco = user;
});
loginViewModel.getPassword().observe(this, user -> {
String coco = user;
});
loginViewModel.getEmail().observe(this, user -> {
String coco = user;
});
}我在xml文件中有引用：
<android.support.design.widget.TextInputLayout
android:id="@+id/userLayout"
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:layout_marginLeft="16dp"
android:layout_marginRight="16dp"
android:layout_marginTop="96dp"
app:layout_constraintEnd_toEndOf="parent"
app:layout_constraintStart_toStartOf="parent"
app:layout_constraintTop_toTopOf="parent">
<android.support.design.widget.TextInputEditText
android:id="@+id/txtUserValue"
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:hint="@string/user"
android:text="@{loginViewModel.user}"
android:inputType="text"/>
</android.support.design.widget.TextInputLayout>
<android.support.design.widget.TextInputLayout
android:id="@+id/passwordLayout"
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:layout_marginLeft="16dp"
android:layout_marginRight="16dp"
android:layout_marginTop="8dp"
app:layout_constraintEnd_toEndOf="parent"
app:layout_constraintStart_toStartOf="parent"
app:layout_constraintTop_toBottomOf="@+id/userLayout">
<android.support.design.widget.TextInputEditText
android:id="@+id/txtPasswordValue"
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:hint="@string/password"
android:text="@{loginViewModel.password}"
android:inputType="textPassword"/>
</android.support.design.widget.TextInputLayout>
<android.support.design.widget.TextInputLayout
android:id="@+id/emailLayout"
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:layout_marginLeft="16dp"
android:layout_marginRight="16dp"
android:layout_marginTop="8dp"
app:layout_constraintEnd_toEndOf="parent"
app:layout_constraintStart_toStartOf="parent"
app:layout_constraintTop_toBottomOf="@+id/passwordLayout">
<android.support.design.widget.TextInputEditText
android:id="@+id/txtEmailValue"
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:hint="@string/email"
android:text="@{loginViewModel.email}"
android:inputType="textWebEmailAddress"/>
</android.support.design.widget.TextInputLayout>为什么我的观察者被称为nerver？有什么帮助吗？谢谢！！o/
回答 1
Stack Overflow用户
回答已采纳
发布于 2019-03-05 05:32:04
我找到了。这就是为什么：
替换为：
android:text="@{loginViewModel.email}"有了这个：
android:text="@={loginViewModel.email}"页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/54991820
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