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我有这样的SQLite数据库:修改项目
private static final String TAG = "DatabaseHelper";
private static final String TABLE_NAME = "people_table";
private static final String COL1 = "_id";
private static final String COL2 = "Hour";
private static final String COL3 = "Minutes";
private static final String COL4 = "On_Off";
如何CAND修改ON_OFF FOM的价值1 - > 0和0 - > 1伪代码:
public void OnAlarm()
{
SQLiteDatabase db = this.getWritableDatabase();
db.execSQL("UPDATE " + TABLE_NAME + " SET " + COL4 + "=1 ");
}
public void OffAlarm()
{
SQLiteDatabase db = this.getWritableDatabase();
db.execSQL("UPDATE " + TABLE_NAME + " SET " + COL4 + "=0 ");
}
我编辑,对不起你的代码不能正常工作 –
我的代码完美的作品。请学习一些基本的SQL语法。 –
android.database.sqlite.SQLiteException:没有这样的列:On_Off(code 1):,while compiling:UPDATE people_table SET On_Off = 0 WHERE On_Off = 1 –