0

我需要帮助将列与字符串进行比较。android 数据库中有一个列,它以字符串的形式存储和关闭。我需要将该列与“On”字符串进行比较,如果该列中有任何 On,我需要它返回 true,如果数据库中没有 on,则返回 false,我该怎么做?

我的代码

公共布尔 QuickTime(){

      String myQuery = "SELECT * FROM DATABASE_TABLE_QUICK WHERE ALARM_ID_QUICK LIKE 'On' ";
      Cursor mCursor = databaseConnect.rawQuery(myQuery, null);
      if(mCursor.getCount()>0)  //if at least one record contains 'on'
          return true;
      else
          return false; //no record was found with the word 'on'
  }
4

2 回答 2

0

一个简单(不是最好)的方法是使用Rawquery。所以你所要做的就是:

private boolean containsOn(SQLiteDatabase mSQLiteDatabase){
    String myQuery = "SELECT * FROM MYTABLE WHERE status LIKE 'on' ";
    Cursor mCursor = mSQLiteDatabase.rawQuery(myQuery, null);
    mCursor.
    if(mCursor.getCount()>0)  //if at least one record contains 'on'
        return true;
    else
        return false; //no record was found with the word 'on'
}

您所要做的就是传递您的 sqlLite 数据库,更改表名和列名的状态列。

于 2012-06-15T01:29:59.267 回答
0
sqlite> create table on_c(id int,onoff varchar(3));
sqlite> insert into on_c (id,onoff)values(1,'ON');
sqlite> insert into on_c (id,onoff)values(2,'OFF');
sqlite> select * from on_c;
1|ON
2|OF
select case count(*) when 0 then 'true' else 'false' end from on_c where onoff like '%N';
false
于 2012-06-15T01:32:23.300 回答