Python - pickle.load()接受一个位置参数(给定2个)

时间:2015-03-02 14:51:16

标签: python pickle

这会产生错误:

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

为什么不腌制数据?

1 个答案:

答案 0 :(得分:3)

这不是您使用pickle.load的方式:

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

从文件反序列化时,它将文件对象作为参数,而不是字符串。

请改为尝试:

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

另一部分(保存)几乎相同,请确保您在范围内有更新的phonebook

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

至于其余代码,您可能想要阅读another answer我给出的非常相似的内容。