Android ViewModel additional arguments

问题:

Is there a way to pass additional argument to my custom AndroidViewModel constructor except Application context. Example:

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);
    }
}

And when I want to user my custom ViewModel class I use this code in my fragment:

MyViewModel myViewModel = ViewModelProvider.of(this).get(MyViewModel.class)

So I don't know how to pass additional argument String param into my custom ViewModel. I can only pass Application context, but not additional arguments. I would really appreciate any help. Thank you.

Edit: I've added some code. I hope it's better now.

回答1:

You need to have a factory class for your ViewModel.

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);
    }
}

And when instantiating the view model, you do like this:

MyViewModel myViewModel = ViewModelProviders.of(this, new MyViewModelFactory(this.getApplication(), "my awesome param")).get(MyViewModel.class);

回答2:

For one factory shared between multiple different view models I'd extend mlyko's answer like this:

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);
        }
    }
}

And instantiating view models:

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);

With different view models having different constructors.

回答3:

I wrote a library that should make doing this more straightforward and way cleaner, no multibindings or factory boilerplate needed, while working seamlessly with ViewModel arguments that can be provided as dependencies by Dagger: 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"
    }    
}

In the view:

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
        })
    }
}

回答4:

I made it a class in which the already created object is passed.

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;
}

And then

ViewModelFactory viewModelFactory = new ViewModelFactory();
viewModelFactory.create(new SampleViewModel(Args1, Args2));
SampleViewModel sampleViewModel = ViewModelProviders.of(this, viewModelFactory).get(SampleViewModel.class);

回答5:

(KOTLIN) My solution uses little bit of Reflection.

Lets say you don't want to create the same looking Factory class every time you create new ViewModel class which needs some arguments. You can accomplish this via Reflection.

For example you would have two different Activities:

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)
    }
}

And ViewModels for those Activities:

class ViewModel1(private val args: Bundle) : ViewModel()

class ViewModel2(private val args: Bundle) : ViewModel()

Then the magic part, Factory class's implementation:

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
        }
    }
}

回答6:

Why not do it like this:

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);
    }
   }
  }
}

and then use it like this in two steps:

MyViewModel myViewModel = ViewModelProvider.of(this).get(MyViewModel.class)
myViewModel.initialize(param)