房间数据库 RawQuery() 不适用于“IN”和“NOT IN”子句

Question

我有一个像 UserTable 这样的表。

@Entity
public class UserTable{
  @PrimaryKey(autoGenerate = true)
    private int userId;
    private String userName;
    private String userEmailId;

  // Below code is getter and setter of this class.
}

@Dao
public interface UserDao {
    @Query("SELECT * FROM userTable")
    public List<UserTable> loadAllUsers();

    @Insert
    public long insertUserTable(UserTable userTable);

    @Insert
    public long[] insertUserTables(UserTable... userTables);

    @Update
    public int updateUserTable(UserTable userTable);

    @Delete
    public int deleteUserTable(UserTable userTable);

    @RawQuery
    public abstract List<UserTable> loadAllUserListByGivenIds
            (SupportSQLiteQuery query);

    public default List<UserTable> loadAllUserListByIds(long[] userIds) {
        List<UserTable> list;
        ArrayList<Object> argsList = new ArrayList<>();
        String selectQuery = "SELECT  * FROM UserTable WHERE userId IN (?);";
        argsList.add(userIds);
        
        SimpleSQLiteQuery simpleSQLiteQuery = new SimpleSQLiteQuery(selectQuery, argsList.toArray());
        list = loadAllUserListByGivenIds(simpleSQLiteQuery);
        return list;
    }
}

// 现在在我的 MainActivity.class 文件中，我使用了以下代码：

 List<UserTable> userList= databaseClient
                .getAppDatabase()
                .userDao()
                .loadAllUserListByIds(new long[]{1L,2L});

我的查询在普通数据库中运行，但是当我传递用户 ID 数组时，在 dao 类的 @RawQuery() 方法中，在 where 条件“WHERE userId IN (?) ”。

如何，我将在房间数据库的@RawQuery() 中使用“IN”子句。

Answer 1

@Query 使用起来要容易得多，它很简单：-

@Query("SELECT * FROM UserTable WHERE userId IN (:idList)")
public List<UserTable> getWhatever(long[] idList);

然后你会使用getWhatever(new long[]{1L,2L})

如果你需要@rawQuery，尽管你可以这样做（为方便起见，使用以前的答案代码）：-

private List<TableXEntity> loadAllUserListByIds(int order,long[] idList) {
    StringBuilder idListAsCSV = new StringBuilder(); //<<<<<<<<<<
    boolean afterFirst = false; //<<<<<<<<<<
    //<<<<<<<<<< all of the loop to create the CSV
    for (Long l: idList) {
        if (afterFirst) {
            idListAsCSV.append(",");
        }
        afterFirst = true;
        idListAsCSV.append(String.valueOf(l));
    }
    StringBuilder sb = new StringBuilder("SELECT * FROM ").append(DBHelper.TableX.NAME);
    sb.append(" WHERE " + DBHelper.TableX.COLUMN_ID + " IN(").append(idListAsCSV).append(") "); //<<<<<<<<<<
    switch (order) {
        case DBHelper.TableX.FIRSTNAME_DESCENDING:
            sb.append(DBHelper.TableX.ORDER_BY_FIRSTNAME_DESC);
            break;
        case DBHelper.TableX.FIRSTNAME_ASCENDING:
            sb.append(DBHelper.TableX.ORDER_BY_FIRSTNAME_ASC);
            break;
        case DBHelper.TableX.LASTNAME_DESCENDING:
            sb.append(DBHelper.TableX.ORDER_BY_LASTNAME_DESC);
            break;
        case DBHelper.TableX.LASTNAME_ASCENDING:
            sb.append(DBHelper.TableX.ORDER_BY_LASTNAME_ASC);
            break;
        default:
            break;
    }
    sb.append(";");
    return roomDao.rawq(new SimpleSQLiteQuery(sb.toString(),null));
}

即提供一个 CSV （虽然我隐约记得能够传递一个数组）

---

要使用绑定参数（作为绑定参数的推荐方法可以防止 SQL 注入），那么您需要一个 ?对于每个值和相应的对象数组。

因此，对于 3 个 id，您需要 IN(?,?,?) 和 Object[] 中的实际值、绑定参数。以下是执行此操作的示例，注意它显示了构建 Object[]（绑定参数/值）的 2 种方法：-

private List<TableXEntity> loadByidList(long[] idlist) {
    List<Object> bindargs = new ArrayList<>(); // way 1
    Object[] args4Bind = new Object[idlist.length]; // way 2
    StringBuilder placeholders = new StringBuilder(); // for the ? placeholders
    /* Build the sql before the place holders */
    StringBuilder sql = new StringBuilder("SELECT * FROM ")
            .append(DBHelper.TableX.NAME)
            .append(" WHERE ")
            .append(DBHelper.TableX.COLUMN_ID)
            .append(" IN (");
    boolean afterfirst = false;
    int i = 0; /* using for each so have index counter (as opposed to for(int i=0 ....) */
    for (long l: idlist) {
        bindargs.add(l); // for way 1
        args4Bind[i++] = String.valueOf(l); // for way 2
        if (afterfirst) {
            placeholders.append(",");
        }
        afterfirst = true;
        placeholders.append("?");
    }
    /* finalise the SQL */
    sql.append(placeholders.toString())
            .append(");");
    //return roomDao.rawq(new SimpleSQLiteQuery(sql.toString(),bindargs.toArray())); // way 1
    return roomDao.rawq(new SimpleSQLiteQuery(sql.toString(),args4Bind)); // way 2
}

Answer 2

请试试这个，它已经工作了！

试试这个简单的技巧来传递 IN 运算符的参数-

List<Object> argList = new ArrayList<>();
 argList.add("3");
 argList.add("6");

    

然后准备你的原始查询字符串：

注意 - 将您的参数列表大小与“？”匹配尺寸

   String selectQuery = "SELECT * FROM task WHERE id IN (?,?)";

之后将原始查询字符串传递给 SimpleSQLiteQuery-

   SimpleSQLiteQuery rawQuery = new SimpleSQLiteQuery(selectQuery, args.toArray());

然后使用 DAO 获取列表：

List<UserTable> taskList1=DatabaseClient
                        .getInstance(getApplicationContext())
                        .getAppDatabase()
                        .userTableDAO()
                        .getAllList(query);