我有这样的一个表:SQLite的多个地方在同一列
Char1 Char2 Difficulty
--------------------------------------------------
jon sara 1
pablo victor 2
laura patricia 1
marta juanjo 3
marina goku 4
我想选择其中的难度= 1和难度= 2。我试图做的所有行:
"SELECT * FROM " + table + " WHERE " + DatabaseOpenHelper.COLUMN_DIFFICULTY + " LIKE '1%'" + " AND " + DatabaseOpenHelper.COLUMN_DIFFICULTY + " LIKE '2%'"
但它不工作,我是SQLite的新手,所以我会很感激任何解释与答案。谢谢!
由于难度列是INT类型,您为什么使用LIKE? –