我有这个SQLite数据库:
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";
我如何修改On_Off fom 1的值 - > 0和0 - > 1 Pseudocod:
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 ");
}
答案 0 :(得分:0)
... WHERE
... 什么?你错过了这个条件!
public void OnAlarm()
{
// Change On_Off from 0 to 1
SQLiteDatabase db = this.getWritableDatabase();
db.execSQL( "UPDATE " + TABLE_NAME + " SET " + COL4 + " = 1 WHERE " + COL4 + " = 0");
}
public void OffAlarm()
{
// Change On_Off from 1 to 0
SQLiteDatabase db = this.getWritableDatabase();
db.execSQL( "UPDATE " + TABLE_NAME + " SET " + COL4 + " = 0 WHERE " + COL4 + " = 1");
}