【发布时间】:2021-10-27 10:55:37
【问题描述】:
早上好,
所以我尝试使用扫描方法创建一个帐户类,该方法对会员卡进行“虚构”扫描,但是,当我在输入字段中输入有效的 card_number 时,它会返回另一个输入字段,但如果我输入一个无效的 card_number 它会返回 Invalid ID
这是我的代码,请问我哪里出错了?
from enum import Enum
from abc import ABC, abstractmethod
import csv
import json
from datetime import datetime, timedelta, date
membersJsonPath = 'members.json'
class BookStatus(Enum):
AVAILABLE, RESERVED, LOANED, LOST = 1, 2, 3, 4
class ReservationStatus(Enum):
WAITING, PENDING, CANCELED, NONE = 1, 2, 3, 4
class AccountStatus(Enum):
ACTIVE, CLOSED, CANCELED, NONE = 1, 2, 3, 4
class Account(ABC):
def __init__(self, status=AccountStatus.ACTIVE):
self.__id = self.scan_id()
self.__status = status
def scan_id(self):
with open(membersJsonPath, 'r') as membersJson:
data = json.load(membersJson)
input_id = input("Please enter your member ID: ")
for member in data['members']:
if member['card_number'] == input_id:
return ("Welcome, " + member['first_name'] + "!")
else:
return "Invalid ID"
a = Account()
a.scan_id()
这是我从 CSV 文件创建的 json 文件的一部分
{
"members": [
{
"\u00ef\u00bb\u00bfid": "1",
"first_name": "Adelaide",
"last_name": "Cunningham",
"gender": "Female",
"email": "a.cunningham@randatmail.com",
"card_number": "13"
},
{
"\u00ef\u00bb\u00bfid": "2",
"first_name": "Charlie",
"last_name": "Roberts",
"gender": "Male",
"email": "c.roberts@randatmail.com",
"card_number": "22"
},
{
"\u00ef\u00bb\u00bfid": "3",
"first_name": "Eric",
"last_name": "Cooper",
"gender": "Male",
"email": "e.cooper@randatmail.com",
"card_number": "33"
},
{
"\u00ef\u00bb\u00bfid": "4",
"first_name": "Cadie",
"last_name": "Hall",
"gender": "Female",
"email": "c.hall@randatmail.com",
"card_number": "43"
}
]
}
【问题讨论】:
-
您的
__init__方法正在调用self.scan_id(),然后在您使用a.scan_id()实例化后再次调用它。 -
我把它改成了 self.__id = None 谢谢。
标签: python json file class methods