我如何可以DRY Android改装电话

Question

我的代码中有几个存储库类。

例如，这是UserRepository:

public class UserRepository {
public static String TAG = "UserRepository";
ApiService mApiService;
SharedPreferences mPrefs;
Context mContext;
RemoteDataSource<User> mRemoteDataSource;

public UserRepository() {
mApiService = new RetrofitClient().getApiService();
mContext = App.getAppContext();
mPrefs = PreferenceManager.getDefaultSharedPreferences(mContext);
mRemoteDataSource = new RemoteDataSource<>();
}
public RemoteDataSource getRemoteDataSource() {
mRemoteDataSource.setIsLoading();
Call<ApiResponse> userCall = mApiService.getUserInfo(mPrefs.getString(User.TOKEN_NAME, null));
userCall.enqueue(new Callback<ApiResponse>() {
@Override
public void onResponse(Call<ApiResponse> call, Response<ApiResponse> response) {
mRemoteDataSource.setIsLoaded();
mRemoteDataSource.setData(response.body().getUser());
mRemoteDataSource.setStatus(response.body().getStatus());
mRemoteDataSource.setMessage(response.body().getMessage());
}
@Override
public void onFailure(Call<ApiResponse> call, Throwable t) {
Log.e(TAG, t.getMessage());
mRemoteDataSource.setFailed(t.getMessage());
}
});
return mRemoteDataSource;
}
}

这是BonusRepository:

public class BonusRepository {
public static String TAG = "BonusRepository";
ApiService mApiService;
SharedPreferences mPrefs;
Context mContext;
LiveData<Bonus> mBonus;
String mId;
RemoteDataSource<Bonus> mRemoteDataSource;
public BonusRepository(String id) {
mId = id;
mApiService = new RetrofitClient().getApiService();
mContext = App.getAppContext();
mPrefs = PreferenceManager.getDefaultSharedPreferences(mContext);
mRemoteDataSource = new RemoteDataSource<>();
}
public RemoteDataSource getRemoteDataSource() {
mRemoteDataSource.setIsLoading();
Call<ApiResponse> bonusCall = mApiService.getBonus(mPrefs.getString(User.TOKEN_NAME, null), mId);
bonusCall.enqueue(new Callback<ApiResponse>() {
@Override
public void onResponse(Call<ApiResponse> call, Response<ApiResponse> response) {
mRemoteDataSource.setIsLoaded();
mRemoteDataSource.setData(response.body().getBonus());
mRemoteDataSource.setStatus(response.body().getStatus());
mRemoteDataSource.setMessage(response.body().getMessage());
}
@Override
public void onFailure(Call<ApiResponse> call, Throwable t) {
Log.e(TAG, t.getMessage());
mRemoteDataSource.setFailed(t.getMessage());
}
});
return mRemoteDataSource;
}
}

除了之外，两个类中的getRemoteDataSource方法是相等的

Call<ApiResponse> userCall = mApiService.getUserInfo(mPrefs.getString(User.TOKEN_NAME, null));

和UserRepository 中的CCD_ 2

不同于：

Call<ApiResponse> bonusCall = ApiService.getBonus(mPrefs.getString(User.TOKEN_NAME, null), mId);

和BonusRepository中的CCD_ 3。

在其他存储库中，我有类似的重复代码。

我想去掉这些重复的东西，但找不到任何好的解决办法。

干燥代码的最佳方法是什么？

Answer 1

为所有存储库创建一个抽象类父级，并实现getRemoteDataSource((方法，在每个特定实现中调用一个新的抽象方法，这将是唯一特定的方法。

例如：

public class AbstractRepository {
protected abstract Call<ApiResponse> performCall();
public RemoteDataSource getRemoteDataSource() {
mRemoteDataSource.setIsLoading();
Call<ApiResponse> userCall = performCall();
userCall.enqueue(new Callback<ApiResponse>() {
@Override
public void onResponse(Call<ApiResponse> call, Response<ApiResponse> response) {
mRemoteDataSource.setIsLoaded();
mRemoteDataSource.setData(response.body().getUser());
mRemoteDataSource.setStatus(response.body().getStatus());
mRemoteDataSource.setMessage(response.body().getMessage());
}
@Override
public void onFailure(Call<ApiResponse> call, Throwable t) {
Log.e(TAG, t.getMessage());
mRemoteDataSource.setFailed(t.getMessage());
}
});
return mRemoteDataSource;
}
}

然后你可以执行类似的操作：

public class UserRepository extends AbstractRepository {
public static String TAG = "UserRepository";
ApiService mApiService;
SharedPreferences mPrefs;
Context mContext;
RemoteDataSource<User> mRemoteDataSource;
public UserRepository() {
mApiService = new RetrofitClient().getApiService();
mContext = App.getAppContext();
mPrefs = PreferenceManager.getDefaultSharedPreferences(mContext);
mRemoteDataSource = new RemoteDataSource<>();
}
protected Call<ApiResponse> performCall() {
return mApiService.getUserInfo(mPrefs.getString(User.TOKEN_NAME, null));
}

}

我让你适应你的需求，但这是最好的解决方案。