2017-08-04 71 views
-2

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 "); 

} 

回答

0

... WHERE ... 什么?你错过了病情!

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"); 

} 
+0

我编辑,对不起你的代码不能正常工作 –

+0

我的代码完美的作品。请学习一些基本的SQL语法。 –

+0

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 –