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Question

我有這樣的SQLite數據庫：

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的價值1 - > 0和0 - > 1僞代碼：

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 "); 

} 

Answer 1

... WHERE ... 什麼？你錯過了病情！

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"); 

}