151

有没有办法将附加参数传递给我的自定义AndroidViewModel构造函数，除了应用程序上下文。例子：

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传递String param给我的 custom ViewModel。我只能传递应用程序上下文，但不能传递其他参数。我真的很感激任何帮助。谢谢你。

编辑：我添加了一些代码。我希望现在好多了。

4

9 回答 9

296

您需要为您的 ViewModel 提供一个工厂类。

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，您可以这样做：

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

对于 kotlin，您可以使用委托属性：

val viewModel: MyViewModel by viewModels { MyViewModelFactory(getApplication(), "my awesome param") }

还有另一个新选项 - 使用工厂的实例化来实现HasDefaultViewModelProviderFactory和覆盖getDefaultViewModelProviderFactory()，然后您可以调用ViewModelProvider(this)或by viewModels()不调用工厂。

于 2017-10-12T08:18:07.703 回答
43

使用依赖注入实现

这对于生产代码来说更先进更好。

Dagger2，Square 的AssistedInject为 ViewModel 提供了生产就绪的实现，可以注入必要的组件，例如处理网络和数据库请求的存储库。它还允许在活动/片段中手动注入参数/参数。以下是基于 Gabor Varadi 的详细文章Dagger Tips使用代码 Gists实现的步骤的简明概述。

Dagger Hilt是下一代解决方案，截至 2020 年 7 月 12 日的 alpha 版本，一旦库处于发布状态，提供相同的用例和更简单的设置。

在 Kotlin 中使用Lifecycle 2.2.0实现

传递参数/参数

// Override ViewModelProvider.NewInstanceFactory to create the ViewModel (VM).
class SomeViewModelFactory(private val someString: String): ViewModelProvider.NewInstanceFactory() {
    override fun <T : ViewModel?> create(modelClass: Class<T>): T = SomeViewModel(someString) as T
} 

class SomeViewModel(private val someString: String) : ViewModel() {
    init {
        //TODO: Use 'someString' to init process when VM is created. i.e. Get data request.
    }
}

class Fragment: Fragment() {
    // Create VM in activity/fragment with VM factory.
    val someViewModel: SomeViewModel by viewModels { SomeViewModelFactory("someString") } 
}

使用参数/参数启用 SavedState

class SomeViewModelFactory(
    private val owner: SavedStateRegistryOwner,
    private val someString: String) : AbstractSavedStateViewModelFactory(owner, null) {
    override fun <T : ViewModel?> create(key: String, modelClass: Class<T>, state: SavedStateHandle) =
            SomeViewModel(state, someString) as T
}

class SomeViewModel(private val state: SavedStateHandle, private val someString: String) : ViewModel() {
    val feedPosition = state.get<Int>(FEED_POSITION_KEY).let { position ->
        if (position == null) 0 else position
    }
        
    init {
        //TODO: Use 'someString' to init process when VM is created. i.e. Get data request.
    }
        
     fun saveFeedPosition(position: Int) {
        state.set(FEED_POSITION_KEY, position)
    }
}

class Fragment: Fragment() {
    // Create VM in activity/fragment with VM factory.
    val someViewModel: SomeViewModel by viewModels { SomeViewModelFactory(this, "someString") } 
    private var feedPosition: Int = 0
     
    override fun onSaveInstanceState(outState: Bundle) {
        super.onSaveInstanceState(outState)
        someViewModel.saveFeedPosition((contentRecyclerView.layoutManager as LinearLayoutManager)
                .findFirstVisibleItemPosition())
    }    
        
    override fun onViewStateRestored(savedInstanceState: Bundle?) {
        super.onViewStateRestored(savedInstanceState)
        feedPosition = someViewModel.feedPosition
    }
}
于 2020-02-08T19:56:04.700 回答
17

对于在多个不同视图模型之间共享的一个工厂，我会像这样扩展 mlyko 的答案：

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);

不同的视图模型具有不同的构造函数。

于 2018-05-16T14:53:27.957 回答
8

基于@vilpe89 以上针对AndroidViewModel 案例的Kotlin 解决方案

class ExtraParamsViewModelFactory(
    private val application: Application,
    private val myExtraParam: String
): ViewModelProvider.NewInstanceFactory() {
    override fun <T : ViewModel?> create(modelClass: Class<T>): T = 
            SomeViewModel(application, myExtraParam) as T
}

然后一个片段可以启动 viewModel 为

class SomeFragment : Fragment() {
    
    // ...

    private val myViewModel: SomeViewModel by viewModels {
        ExtraParamsViewModelFactory(this.requireActivity().application, "some string value")
    }

    // ...

}

然后是实际的 ViewModel 类

class SomeViewModel(application: Application, val myExtraParam:String) : AndroidViewModel(application) {
    // ...
}

或以某种合适的方法...

override fun onActivityCreated(...){
    // ...
    val myViewModel = ViewModelProvider(this, ExtraParamsViewModelFactory(this.requireActivity().application, "some string value")).get(SomeViewModel::class.java)
    // ...
}
于 2020-05-05T11:25:06.640 回答
4

我将它设为一个类，其中传递了已创建的对象。

private Map<String, ViewModel> viewModelMap;

public ViewModelFactory() {
    this.viewModelMap = new HashMap<>();
}

public void add(ViewModel viewModel) {
    viewModelMap.put(viewModel.getClass().getCanonicalName(), viewModel);
}

@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.getCanonicalName())) {
            return (T) viewModel.getValue();
        }
    }
    return null;
}

接着

ViewModelFactory viewModelFactory = new ViewModelFactory();
viewModelFactory.add(new SampleViewModel(arg1, arg2));
SampleViewModel sampleViewModel = ViewModelProviders.of(this, viewModelFactory).get(SampleViewModel.class);
于 2019-07-05T18:15:10.710 回答
2 
class UserViewModelFactory(private val context: Context) : ViewModelProvider.NewInstanceFactory() {
 
    override fun <T : ViewModel?> create(modelClass: Class<T>): T {
        return UserViewModel(context) as T
    }
 
}

class UserViewModel(private val context: Context) : ViewModel() {
 
    private var listData = MutableLiveData<ArrayList<User>>()
 
    init{
        val userRepository : UserRepository by lazy {
            UserRepository
        }
        if(context.isInternetAvailable()) {
            listData = userRepository.getMutableLiveData(context)
        }
    }
 
    fun getData() : MutableLiveData<ArrayList<User>>{
        return listData
    }

在 Activity 中调用 Viewmodel

val userViewModel = ViewModelProviders.of(this,UserViewModelFactory(this)).get(UserViewModel::class.java)

更多参考：Android MVVM Kotlin 示例

于 2020-07-11T11:20:14.750 回答
0

我编写了一个库，它应该使这样做更简单、更简洁，不需要多重绑定或工厂样板，同时与 Dagger 可以作为依赖项提供的 ViewModel 参数无缝工作： 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
        })
    }
}
于 2018-11-21T19:15:21.750 回答
0

(KOTLIN) 我的解决方案使用了一点点反射。

假设您不想每次创建需要一些参数的新 ViewModel 类时都创建相同的 Factory 类。您可以通过反射来完成此操作。

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)
    }
}

以及这些活动的 ViewModel：

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

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

然后是神奇的部分，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
        }
    }
}
于 2019-02-13T10:22:02.960 回答
-1

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)
于 2019-03-03T03:57:04.247 回答