java - 合并多个 RealmList 并对结果列表进行排序？

Question

我刚开始在我当前的 Android 应用程序中使用 Realm，到目前为止它很棒。不幸的是我遇到了一个问题:

在我的应用程序中，用户可以在他的日记中添加不同类型的条目(他今天吃了什么？他喝了什么饮料？等等)。一个 DiaryEntry 对象表示给定日期(例如 21.05.2017 等)的所有条目的总和。

public class DiaryEntry extends RealmObject {

    // ID of the day this diary entry represents
    @PrimaryKey private Integer dateId;

    private RealmList<MealEntry> mealEntries;
    private RealmList<DrinkEntry> drinkEntries;
    private RealmList<SymptomEntry> symptomEntries;
    private RealmList<MedicineEntry> medicineEntries;

    public void addMealEntry(MealEntry mealEntry) {
        mealEntries.add(mealEntry);
    }
    public RealmList<MealEntry> getMealEntries() {
        return mealEntries;
    }

    public void addDrinkEntry(DrinkEntry drinkEntry) {
        drinkEntries.add(drinkEntry);
    }
    public RealmList<DrinkEntry> getDrinkEntries() {
        return drinkEntries;
    }

    public void addSymptomEntry(SymptomEntry symptomEntry) {
        symptomEntries.add(symptomEntry);
    }
    public RealmList<SymptomEntry> getSymptomEntries() {
        return symptomEntries;
    }

    public void addMedicineEntry(MedicineEntry medicineEntry) {
        medicineEntries.add(medicineEntry);
    }
    public RealmList<MedicineEntry> getMedicineEntries() {
        return medicineEntries;
    }
}

要在日记中显示特定日期的数据，所有条目都应按用户创建它们的时间排序。所以每个条目对象都包含一个字段“时间”。

private int time;

我想出了一个临时解决方案来解决我的问题，但它远非完美。以下代码是在 UI Thread 上执行的，这显然是一种不好的做法。

List<RealmObject> entryList = new ArrayList<>();
        OrderedRealmCollectionSnapshot<MealEntry> mealEntries = diaryEntry.getMealEntries().createSnapshot();
        OrderedRealmCollectionSnapshot<DrinkEntry> drinkEntries = diaryEntry.getDrinkEntries().createSnapshot();
        OrderedRealmCollectionSnapshot<MedicineEntry> medicineEntries = diaryEntry.getMedicineEntries().createSnapshot();
        OrderedRealmCollectionSnapshot<SymptomEntry> symptomEntries = diaryEntry.getSymptomEntries().createSnapshot();

        entryList.addAll(mealEntries);
        entryList.addAll(drinkEntries);
        entryList.addAll(medicineEntries);
        entryList.addAll(symptomEntries);

        Collections.sort(entryList, entryComparator);

排序条目列表的代码通过调用 time 字段的 getter 方法使用反射:

public int compare(RealmObject entry1, RealmObject entry2) {
            try {
                Method timeGetter1 = entry1.getClass().getMethod("getTime");
                Method timeGetter2 = entry2.getClass().getMethod("getTime");

                int time1 = (Integer) timeGetter1.invoke(entry1);
                int time2 = (Integer) timeGetter2.invoke(entry2);

                return time1 - time2;
            } catch (NoSuchMethodException e) {
                e.printStackTrace();
                Timber.d("No such method 'getTime'.");
           }
      // Other catch clauses

正如我之前所说，所有这些都发生在 UI 线程上。

我知道我不能跨线程传递 RealmObject、RealmList 和 RealmResult 所以我真的很难来为此提出一个异步解决方案。我想启动一个后台线程，并在其中创建一个 DiaryEntry 对象中所有 RealmList 的副本。然后合并这个非托管列表并对其进行排序 - 所有这些都在后台线程上进行。

所以我的问题是:是否有任何首选策略来合并多个 RealmList 并对合并后的列表进行排序 - 所有这些都以异步方式进行？我上面描述的尝试是否有效？

Answer 1

感谢@EpicPandaForce

我完全按照你描述的方式解决了它，它就像一个魅力 - 现在我什至拥有实时功能，不需要手动刷新数据，很好 :)

如果有人遇到同样的问题，我会在这里发布一些代码 fragment ，展示我是如何用代码解决它的。

public class Entry extends RealmObject {
    private static final int ENTRY_MEAL = 0;
    private static final int ENTRY_DRINK = 1;
    private static final int ENTRY_SYMPTOM = 2;
    private static final int ENTRY_MEDICINE = 3;

    /** The tag describes what kind of entry it represents */
    private int tag;

    /* Only one of these can be set, according to what this entry represents. */
    @Nullable private MealEntry mealEntry;
    @Nullable private DrinkEntry drinkEntry;
    @Nullable private SymptomEntry symptomEntry;
    @Nullable private MedicineEntry medicineEntry;

    /** The time value this entry was created at */
    /** Format: hours + minutes * 60 */
    private int time;

    public int getTime() {
        return time;
    }

/* Can only be accessed from within the 'data' package */

    void setTime(int time) {
        this.time = time;
    }

/**
     * Creates a new entry object in the realm database and tags it as 'MEAL'
     *
     * @param realm not null
     * @param mealEntry the {@link MealEntry} object to map this entry to, not null
     *
     * @return the newly created entry
     */
    static Entry createEntryAsMeal(@NonNull final Realm realm, @NonNull final MealEntry mealEntry) {
        if(realm == null) {
            throw new IllegalArgumentException("'realm' may not be null");
        }
        if(mealEntry == null) {
            throw new IllegalArgumentException("'mealEntry' may not be null");
        }

        Entry entry = realm.createObject(Entry.class);
        entry.tag = ENTRY_MEAL;
        entry.mealEntry = mealEntry;
        return entry;
        }

/* Same methods for other tag types ... */

在 MealEntry.class 中:

public class MealEntry extends RealmObject {

    @PrimaryKey @Required private String id;

    @Required private String title;

    /** The entry objects this meal-entry is added to */
    Entry entry;

    /** This time value describes when the user consumed this meal **/
    private int time;
// other fields

/**
* Creates a new MealEntry object in the realm.
 * <p>
 *     Note: It is important to use this factory method for creating {@link MealEntry} objects in realm.
 *     Under the hood, a {@link Entry} object is created for every MealEntry and linked to it.
 * </p>
 *
 * @param realm not null
 *
 * @return new MealEntry object which has been added to the <code>realm</code>
 */
public static MealEntry createInRealm(@NonNull Realm realm) {
    if(realm == null) {
        throw new IllegalArgumentException("'realm' may not be null");
    }

    MealEntry mealEntry = realm.createObject(MealEntry.class, UUID.randomUUID().toString());
    mealEntry.entry = Entry.createEntryAsMeal(realm, mealEntry);
    return mealEntry;
}

“时间”字段存在于 Entry.class 和 MealEntry.class 中，因此如果后者发生变化，则必须相应地更新 Entry:

/**
     * Sets the time value for the <code>mealEntry</code> to the specified value.
     * <p>
     *     Note: This method is necessary in order to sync the new time value with the underlying
     *     {@link Entry} object that is connected with the <code>mealEntry</code>.
     * </p>
     *
     * @param mealEntry the {@link MealEntry} object to set the time for, not null
     *
     * @param time the new time value, must be in range of [0, 24*60] because of the format: hours*60 + minutes
     *
     */
    public static void setTimeForMealEntry(@NonNull MealEntry mealEntry, @IntRange(from=0, to=24*60) int time) {
        if(mealEntry == null) {
            throw new IllegalArgumentException("'mealEntry' may not be null");
        }

        mealEntry.setTime(time);

        Entry entry = mealEntry.entry;
        if(entry == null) {
            throw new IllegalStateException("'mealEntry' contains no object of type 'Entry'! Something went wrong on creation of the 'mealEntry'");
        }

        /* Syncs the entries time value with the time value for this MealEntry. */
        /* That´s important for sorting a list of all entries. */
        entry.setTime(time);
    }

注意:我可以只在 MealEntry 中存储相应 Entry 对象的 ID，反之亦然，Entry 对象将 ID 存储到相应的 MealEntry 对象。但是我不知道这在性能上有什么不同，所以我只是采用了上述方法。另一种方法的一个原因是我不必将“时间”字段存储两次，一次在 Entry.class 中，一次在 MealEntry.class 中，因为在 Entry.class 中我可以获得时间通过ID找到对应的MealEntry对象，然后获取时间。