Python - pickle.load() 采用一个位置参数(给定 2 个)

Question

这会产生一个错误:

pickle.load() takes one positional argument (2 given)

这是我的代码:

import pickle, os.path

created = False

phoneBook = {}
name = input("Please enter a name(or press enter to end input): ")
while name != '':
    number = input("Please enter number: ")
    phoneBook[name] = number
    name = input("Please enter a name(or press enter to end input): ")
    if name == '':
        print("Thank You!")

print("Your phonebook contains the following entries:")
for name, number in phoneBook.items():
    print("%s - %s" % (name, number))

while not created:
    if not os.path.isfile('phonebook.json'):
        phoneBook_Ori = pickle.load('phonebook.json', 'r')
        created = True
    else:
        phoneBook_Ori = pickle.load('phonebook.json', 'w')
        phoneBook_Upd = phoneBook_Ori.update(phoneBook)
        phoneBook_Ori.write(phoneBook_Upd)

phoneBook_Ori.close

为什么不 pickle 数据？

Answer 1

这不是你使用pickle.load的方式:

phoneBook_Ori = pickle.load('phonebook.json', 'r')

从文件反序列化时，它采用文件对象作为参数，而不是字符串。

试试这个:

# create file object with permissions
with open('phonebook.json', 'r') as f:
    # load using pickle de-serializer
    phoneBook_Ori = pickle.load(f)

保存几乎相同，请确保您的范围内有更新的电话簿:

with open('phonebook.json', 'wb') as f:
    phoneBook_Ori = pickle.dump(phonebook, f)

至于其余代码，您可能需要阅读another answer我已经给出了非常相似的信息。