Question

我正在RoomDatabase上执行SQL INSERT语句。使用RoomDatabase.query()进行相同操作。但是，除非我在返回的cursor上调用任何函数，否则插入的行不会提交到数据库 -

@Entity
public class Person {
    @PrimaryKey
    @NonNull
    private String uuid;

    @Embedded
    @NonNull
    private Address address;

    public Person(@NonNull String uuid, @NonNull Address address) {
        this.uuid = uuid;
        this.address = address;
    }

    @NonNull
    public String getUuid() {
        return uuid;
    }

    @NonNull
    public Address getAddress() {
        return address;
    }
}


public class Address {
    @NonNull
    private String street;

    @NonNull
    private String city;

    public Address(String street, String city) {
        this.street = street;
        this.city = city;
    }

    @NonNull
    public String getStreet() {
        return street;
    }

    @NonNull
    public String getCity() {
        return city;
    }
}

@Dao
public interface PersonDao {
    @Query("SELECT * FROM Person")
    List<Person> getPersons();
}

@TypeConverters(DateConverter.class)
@Database(entities = {Person.class}, version = 1, exportSchema = false)
public abstract class PersonDb extends RoomDatabase {
    private static volatile PersonDb INSTANCE;

    public abstract PersonDao getPersonDao();

    public static PersonDb getInstance(Context context) {
        if (INSTANCE == null) {
            synchronized (PersonDb.class) {
                if (INSTANCE == null) {
                    INSTANCE = Room.databaseBuilder(context.getApplicationContext(),
                            PersonDb.class, "persons.db")
                            .allowMainThreadQueries()
                            .build();
                }
            }
        }
        return INSTANCE;
    }
}

这是我的测试用例 -

@RunWith(AndroidJUnit4.class)
public class PersonsIssueTest {
    private final static String TAG = PersonsIssueTest.class.getSimpleName();

    private PersonDb mDb;

    private SimpleSQLiteQuery personsInsertQuery =
            new SimpleSQLiteQuery("INSERT INTO Person (uuid, street, city) " +
                    "VALUES ('1', 'lane', 'york')");

    @After
    public void closeDb() throws IOException {
        mDb.close();
    }

    @Test
    public void insertQueryTest() {
        Context context = InstrumentationRegistry.getTargetContext();

        Cursor c;

        mDb = PersonDb.getInstance(context);
        PersonDao dao = mDb.getPersonDao();

        List<Person> persons = dao.getPersons();
        Log.d(TAG, "> number of persons " + persons.size() + " rows");

        Log.d(TAG, "> running: " + personsInsertQuery.getSql());
        c = mDb.query(personsInsertQuery);
        /* If I comment the next statement, number of persons returned is 0 in the following dao.getPersons() */
        c.moveToFirst();
        c.close();

        persons = dao.getPersons();
        Log.d(TAG, "> number of persons " + persons.size() + " rows");
        assertEquals(1, persons.size());
    }

}

如果我评论c.moveToFirst()，则以下dao.getPersons()

中返回的人数为0

这里有任何问题吗？

Answer 1

Room提供了一个内置插入方法，您可以按如下方式设置

@Dao
public interface PersonDao {
    @Query("SELECT * FROM Person")
    List<Person> getPersons();

    @Insert
    void insertPersonData(Person person);
}

现在您创建一个人物实例，填充其值并使用

插入

 dao.insertPersonData(personObject);

RoomDatabase.query（），其中INSERT语句不提交对数据库的更改

1 个答案: