有没有办法将其他参数传递给我的自定义AndroidViewModel构造函数，除了Application上下文。例：
public class MyViewModel extends AndroidViewModel {
private final LiveData<List<MyObject>> myObjectList;
private AppDatabase appDatabase;
public MyViewModel(Application application, String param) {
super(application);
appDatabase = AppDatabase.getDatabase(this.getApplication());
myObjectList = appDatabase.myOjectModel().getMyObjectByParam(param);
}
}
当我想使用我的自定义ViewModel类时，我在我的片段中使用此代码：
MyViewModel myViewModel = ViewModelProvider.of(this).get(MyViewModel.class)
所以我不知道如何将额外的参数String param传递到我的自定义ViewModel。我只能传递Application上下文，但不能传递其他参数。我真的很感激任何帮助。谢谢。
编辑：我添加了一些代码。我希望现在好多了。
您需要为ViewModel建立工厂类。
public class MyViewModelFactory implements ViewModelProvider.Factory {
private Application mApplication;
private String mParam;
public MyViewModelFactory(Application application, String param) {
mApplication = application;
mParam = param;
}
@Override
public <T extends ViewModel> T create(Class<T> modelClass) {
return (T) new MyViewModel(mApplication, mParam);
}
}
在实例化视图模型时，您会这样做：
MyViewModel myViewModel = ViewModelProviders.of(this, new MyViewModelFactory(this.getApplication(), "my awesome param")).get(MyViewModel.class);
对于在多个不同视图模型之间共享的一个工厂，我将扩展mlyko的答案，如下所示：
public class MyViewModelFactory extends ViewModelProvider.NewInstanceFactory {
private Application mApplication;
private Object[] mParams;
public MyViewModelFactory(Application application, Object... params) {
mApplication = application;
mParams = params;
}
@Override
public <T extends ViewModel> T create(Class<T> modelClass) {
if (modelClass == ViewModel1.class) {
return (T) new ViewModel1(mApplication, (String) mParams[0]);
} else if (modelClass == ViewModel2.class) {
return (T) new ViewModel2(mApplication, (Integer) mParams[0]);
} else if (modelClass == ViewModel3.class) {
return (T) new ViewModel3(mApplication, (Integer) mParams[0], (String) mParams[1]);
} else {
return super.create(modelClass);
}
}
}
并实例化视图模型：
ViewModel1 vm1 = ViewModelProviders.of(this, new MyViewModelFactory(getApplication(), "something")).get(ViewModel1.class);
ViewModel2 vm2 = ViewModelProviders.of(this, new MyViewModelFactory(getApplication(), 123)).get(ViewModel2.class);
ViewModel3 vm3 = ViewModelProviders.of(this, new MyViewModelFactory(getApplication(), 123, "something")).get(ViewModel3.class);
具有不同构造函数的不同视图模型。
我写了一个库，应该使这个更简单，更清洁，不需要多绑定或工厂样板，同时与ViewModel参数无缝协作，可以作为Dagger的依赖项提供：https://github.com/radutopor/ViewModelFactory
@ViewModelFactory
class UserViewModel(@Provided repository: Repository, userId: Int) : ViewModel() {
val greeting = MutableLiveData<String>()
init {
val user = repository.getUser(userId)
greeting.value = "Hello, $user.name"
}
}
在视图中：
class UserActivity : AppCompatActivity() {
@Inject
lateinit var userViewModelFactory2: UserViewModelFactory2
override fun onCreate(savedInstanceState: Bundle?) {
super.onCreate(savedInstanceState)
setContentView(R.layout.activity_user)
appComponent.inject(this)
val userId = intent.getIntExtra("USER_ID", -1)
val viewModel = ViewModelProviders.of(this, userViewModelFactory2.create(userId))
.get(UserViewModel::class.java)
viewModel.greeting.observe(this, Observer { greetingText ->
greetingTextView.text = greetingText
})
}
}
（KOTLIN）我的解决方案使用了一点点反射。
让我们假设您不希望每次创建需要一些参数的新ViewModel类时都创建相同的Factory类。你可以通过Reflection完成这个。
例如，您将有两个不同的活动：
class Activity1 : FragmentActivity() {
override fun onCreate(savedInstanceState: Bundle?) {
super.onCreate(savedInstanceState)
val args = Bundle().apply { putString("NAME_KEY", "Vilpe89") }
val viewModel = ViewModelProviders.of(this, ViewModelWithArgumentsFactory(args))
.get(ViewModel1::class.java)
}
}
class Activity2 : FragmentActivity() {
override fun onCreate(savedInstanceState: Bundle?) {
super.onCreate(savedInstanceState)
val args = Bundle().apply { putInt("AGE_KEY", 29) }
val viewModel = ViewModelProviders.of(this, ViewModelWithArgumentsFactory(args))
.get(ViewModel2::class.java)
}
}
和ViewModels用于这些活动：
class ViewModel1(private val args: Bundle) : ViewModel()
class ViewModel2(private val args: Bundle) : ViewModel()
然后神奇的部分，Factory类的实现：
class ViewModelWithArgumentsFactory(private val args: Bundle) : NewInstanceFactory() {
override fun <T : ViewModel?> create(modelClass: Class<T>): T {
try {
val constructor: Constructor<T> = modelClass.getDeclaredConstructor(Bundle::class.java)
return constructor.newInstance(args)
} catch (e: Exception) {
Timber.e(e, "Could not create new instance of class %s", modelClass.canonicalName)
throw e
}
}
}
我把它作为一个已经创建的对象传递的类。
private Map<String, ViewModel> viewModelMap;
public ViewModelFactory() {
this.viewModelMap = new HashMap<>();
}
public void create(ViewModel viewModel) {
viewModelMap.put(viewModel.getClass().getSimpleName(), viewModel);
create(viewModel.getClass());
}
@NonNull
@Override
public <T extends ViewModel> T create(@NonNull Class<T> modelClass) {
for (Map.Entry<String, ViewModel> viewModel : viewModelMap.entrySet()) {
if (viewModel.getKey().equals(modelClass.getSimpleName())) {
return (T) viewModel.getValue();
}
}
return null;
}
然后
ViewModelFactory viewModelFactory = new ViewModelFactory();
viewModelFactory.create(new SampleViewModel(Args1, Args2));
SampleViewModel sampleViewModel = ViewModelProviders.of(this, viewModelFactory).get(SampleViewModel.class);
为什么不这样做：
public class MyViewModel extends AndroidViewModel {
private final LiveData<List<MyObject>> myObjectList;
private AppDatabase appDatabase;
private boolean initialized = false;
public MyViewModel(Application application) {
super(application);
}
public initialize(String param){
synchronized ("justInCase") {
if(! initialized){
initialized = true;
appDatabase = AppDatabase.getDatabase(this.getApplication());
myObjectList = appDatabase.myOjectModel().getMyObjectByParam(param);
}
}
}
}
然后分两步使用它：
MyViewModel myViewModel = ViewModelProvider.of(this).get(MyViewModel.class)
myViewModel.initialize(param)