SQLiteException: bind or column index out of range

来源：互联网发布：死人的衣服在淘宝卖编辑：程序博客网时间：2024/06/10 04:31

今天写一个登录的时候查询数据库时报了这样的错误

出错代码：

private PhoneNumber queryPhone(String phonenumber) {SQLiteDatabase sqLiteDatabase = DB.getWritableDatabase();PhoneNumber phoneNumber = new PhoneNumber();Cursor cursor = sqLiteDatabase.query("PhoneNumber", new String[] {"id", "phonenumber", "passwd", "username" }, "phonenumber",new String[] { phonenumber }, null, null, null);while (cursor.moveToNext()) {phoneNumber.id = cursor.getInt(cursor.getColumnIndex("id"));phoneNumber.phoneNumber = cursor.getString(cursor.getColumnIndex("phonenumber"));phoneNumber.passwd = cursor.getString(cursor.getColumnIndex("passwd"));phoneNumber.username = cursor.getString(cursor.getColumnIndex("username"));}DB.close();sqLiteDatabase.close();return phoneNumber;}

查询条件忘记加“=？”了

改正之后代码：

private PhoneNumber queryPhone(String phonenumber) {SQLiteDatabase sqLiteDatabase = DB.getWritableDatabase();PhoneNumber phoneNumber = new PhoneNumber();Cursor cursor = sqLiteDatabase.query("PhoneNumber", new String[] {"id", "phonenumber", "passwd", "username" }, "phonenumber=?",new String[] { phonenumber }, null, null, null);while (cursor.moveToNext()) {phoneNumber.id = cursor.getInt(cursor.getColumnIndex("id"));phoneNumber.phoneNumber = cursor.getString(cursor.getColumnIndex("phonenumber"));phoneNumber.passwd = cursor.getString(cursor.getColumnIndex("passwd"));phoneNumber.username = cursor.getString(cursor.getColumnIndex("username"));}DB.close();sqLiteDatabase.close();return phoneNumber;}

这样就对了记录下下次不要犯这样的错误