迁移房间数据库时有没有办法使用迁移期间计算的变量？

Question

我正在用 Java 编写一个 Android 应用程序，并且我已经更改了我的数据库，现在我想将它迁移到新版本，但为此我需要计算一些旧值，然后将它们插入其中一个新表进入数据库但是当我调用时

database.execSQL()

使用我的变量然后访问它们似乎没有设置它们，因为每个值都是 0。

这是不可能的，还是我只是错过了什么？

public void migrate(@NonNull SupportSQLiteDatabase database) {
             database.execSQL("CREATE TABLE 'season' ('id' INTEGER PRIMARY KEY AUTOINCREMENT NOT NULL, 
                    'spring' INTEGER DEFAULT 0, 'summer' INTEGER DEFAULT 0, 'autumn' INTEGER 
                    DEFAULT 0, 'winter' 
                    INTEGER DEFAULT 0, 'spook' INTEGER DEFAULT 0)");
            database.execSQL("INSERT INTO 'Clothes' (type_of_clothing,clothing_name,description,seasonId,in_laundry,TypeList) " +
                    "SELECT type_of_clothing,clothing_name,description,type_of_clothing,in_laundry,TypeList " +
                    "FROM ClothesNew");
            Cursor cursor = database.query("SELECT * FROM Clothes");
            Cursor secondCursor = database.query("SELECT * FROM ClothesSeason");
            ArrayList<Season> seasonsToBeAdded = new ArrayList<>();
            while (cursor.moveToNext()) {
                secondCursor.moveToNext();
                int Sid = cursor.getInt(0);
                String seasonString = secondCursor.getString(2);
                Season season = new Season(seasonString);
                int tempId = Sid;
                boolean isInList = false;
                for (Season tempSeason : seasonsToBeAdded) {
                    if (tempSeason.equals(season)) {
                        tempId = seasonsToBeAdded.indexOf(tempSeason);
                        isInList = true;
                        break;
                    }
                }
                if (!isInList) {
                    season.setId(seasonsToBeAdded.size());
                    tempId = season.getId();
                    seasonsToBeAdded.add(season);
                }
                if (seasonsToBeAdded.size() == 0) {seasonsToBeAdded.add(season);}
                
                database.execSQL("UPDATE Clothes SET seasonId = :tempId WHERE uid = :Sid");
            }

            int id, spring, summer, autumn, winter, spook;

            for (Season season : seasonsToBeAdded) {
                id = season.getId();
                spring = season.getSpring();
                summer = season.getSummer();
                autumn = season.getAutumn();
                winter = season.getWinter();
                spook = season.getSpook();
                database.execSQL("INSERT INTO season " +
                        "VALUES(:id, :spring, :summer, :autumn, :winter, :spook) ");
            }
            Cursor otherCursor = database.query("SELECT * FROM season");
            ArrayList<Integer> springValues= new ArrayList<>();
            while (otherCursor.moveToNext()) {
                springValues.add(otherCursor.getInt(1));
            }
        }

当我查看存储在 springValues 中的值时，即使我在此查询中添加了正确的值，它们也都是 0

database.execSQL("INSERT INTO season " +
                 "VALUES(:id, :spring, :summer, :autumn, :winter, :spook) ");

我可以不在这里使用 :var 符号吗？

Answer 1

我建议使用 SQLiteDatabase Insert 便捷方法和 ContentValues 对象。

而不是：-

        for (Season season : seasonsToBeAdded) {
            id = season.getId();
            spring = season.getSpring();
            summer = season.getSummer();
            autumn = season.getAutumn();
            winter = season.getWinter();
            spook = season.getSpook();
            database.execSQL("INSERT INTO season " +
                    "VALUES(:id, :spring, :summer, :autumn, :winter, :spook) ");
        }

您的代码可能是：-

    ContentValues cv = new ContentValues(); // Instantiate ContentValues object
    for (Season season : seasonsToBeAdded) {
        cv.clear();
        cv.put("`id`",season.getId());
        cv.put("`spring`",season.getSpring());
        cv.put("`summer`",season.getSummer());
        cv.put("`autumn`",season.getAutumn());
        cv.put("`winter`",season.getWinter());
        cv.put("`spook`",season.getSpook());
        database.insert("season",null,cv); //<<<< Recommended Uses the convenience Insert Method

• 请注意，列名不必像上面使用的那样用重音 `` 括起来。

• 以上代码为in-principal code，未经测试，可能存在一些错误。

如您所见，SQL 是为您构建的。 insert 方法返回插入行的 rowid（id 列），如果插入失败则返回 -1，您可能希望利用该值。

您可能还希望使用其他插入方法，例如insertWithOnConflict 例如database.insertWithOnConflict("season",null,cv,SQLiteDatabase.CONFLICT_IGNORE);

还建议通过使用适当的 SQLiteDatabase 事务处理方法来利用事务（请参阅beginTransaction）

还建议您对列名使用常量，以便在适当的位置对名称进行编码。列名错误的可能性很小。