java - 找到空游标时如何返回

Question

我有以下代码从 SQLite 数据库读取数据。我可以检测到空光标，但我不知道如何摆脱它。我无法将 return 语句移至 if( result != null && result.moveToFirst() 语句内。我得到最后两行日志然后程序崩溃

/CSV Import﹕ cursor index is NOT greater than 0
/CSV Import﹕ Null Cursor Found

代码:

public Equipment getEquipmentByID(int id){
    SQLiteDatabase db = this.getReadableDatabase();

    String SQL = "SELECT "+EQUIPMENT_TABLE+".*, "+LOCATION_TABLE+"."+LOCATION_NAME+" as locationName FROM "+ EQUIPMENT_TABLE + " " +
            "JOIN "+LOCATION_TABLE + " ON "+LOCATION_TABLE+"."+LOCATION_ID+"="+EQUIPMENT_TABLE+"."+EQUIPMENT_LOCATION+" " +
                    "WHERE "+EQUIPMENT_TABLE+"."+EQUIPMENT_ID +"="+id;

    Cursor result = db.rawQuery(SQL, null);
    result.moveToFirst();

    if(result != null && result.moveToFirst()) {
        Log.v("CSV Import", "Null Cursor not Found");
    } else {
        if (result.getCount()>0) {
            Log.v("CSV Import", "cursor index is grater than 0");
        } else {
            Log.v("CSV Import", "cursor index is NOT greater than 0");
        }
        Log.v("CSV Import", "Null Cursor Found");
    }

    Equipment equipment = new Equipment(result.getInt(result.getColumnIndex(EQUIPMENT_ID)), result.getString(result.getColumnIndex(EQUIPMENT_NAME)),
            result.getString(result.getColumnIndex(EQUIPMENT_MODEL)), result.getString(result.getColumnIndex(EQUIPMENT_SERIAL)),
            result.getString(result.getColumnIndex(EQUIPMENT_NOTES)), new Location(result.getInt(result.getColumnIndex(EQUIPMENT_LOCATION)),
            result.getString(result.getColumnIndex("locationName"))), new EquipmentType(1, "Undefined"));
    return equipment;
}

Answer 1

From the docs , 如果光标为空，moveToFirst() 将返回 false，否则返回 true。因此，从以下行开始: result.moveToFirst(); 全部替换为:

if(result == null) return null;

if(!result.moveToFirst()) {
    result.close();
    db.close();
    return null; 
}

Equipment equipment = new Equipment(result.get //.... and so on ....//);
result.close();
db.close();
return equipment;