以下selectQuery有什么问题吗？

Question

int getIdForSong(Song song){

    String selectQuery = "SELECT id FROM " + TABLE_SONG + " WHERE " + SONG_TITLE + "=" + song.getSongTitle() + " AND " + ARTIST_NAME + "=" + song.getArtistName();
    SQLiteDatabase db = this.getReadableDatabase();
    Cursor cursor = db.rawQuery(selectQuery, null);
    int id = Integer.parseInt(cursor.getString(0));
    return id;
}

我在“=”附近得到和异常。任何人？

Answer 1

将selectQuery Value更改为：

"SELECT id FROM " + TABLE_SONG + " WHERE " + SONG_TITLE + " = '" + song.getSongTitle() + "' AND  '" + ARTIST_NAME + "' = '" + song.getArtistName() + "'";

Answer 2

试试这个

int getIdForSong(Song song){

String selectQuery = "SELECT id FROM " + TABLE_SONG + " WHERE " + SONG_TITLE + "= ' " + song.getSongTitle() + "' AND " + ARTIST_NAME + "= ' " + song.getArtistName()+" ' ";
SQLiteDatabase db = this.getReadableDatabase();
Cursor cursor = db.rawQuery(selectQuery, null);
int id = Integer.parseInt(cursor.getString(0));
return id;
}

Answer 3

对于CursorOutOfBoundException，

Cursor cursor = db.rawQuery（selectQuery，null）之后; ，

if（null！= cursor＆amp;＆amp; cursor.moveToFirst（））{

  int id = Integer.parseInt(cursor.getString(0));
  return id;

}