我是Python的初学者并遇到了错误。我正在尝试创建一个程序,该程序将获取用户创建的用户名和密码,将它们写入列表并将这些列表写入文件。这是我的一些代码: 这是用户创建用户名和密码的部分。
userName=input('Please enter a username')
password=input('Please enter a password')
password2=input('Please re-enter your password')
if password==password2:
print('Your passwords match.')
while password!=password2:
password2=input('Sorry. Your passwords did not match. Please try again')
if password==password2:
print('Your passwords match')
我的代码工作正常,直到此时我收到错误:
无效文件:< _io.TextIOWrapper name ='usernameList.txt'mode ='wt'coding ='cp1252'>。
我不确定为什么会返回此错误。
if password==password2:
usernames=[]
usernameFile=open('usernameList.txt', 'wt')
with open(usernameFile, 'wb') as f:
pickle.dump(usernames,f)
userNames.append(userName)
usernameFile.close()
passwords=[]
passwordFile=open('passwordList.txt', 'wt')
with open(passwordFile, 'wb') as f:
pickle.dump(passwords,f)
passwords.append(password)
passwordFile.close()
有没有办法修复错误,或者将列表写入文件的其他方法? 感谢
答案 0 :(得分:1)
你有正确的想法,但有很多问题。当用户密码不匹配时,通常会再次提示两者。
with
块旨在打开和关闭您的文件,因此无需在末尾添加close
。
下面的脚本显示了我的意思,然后您将拥有两个持有Python list
的文件。因此,尝试查看它没有多大意义,您现在需要将相应的读取部分写入您的代码。
import pickle
userName = input('Please enter a username: ')
while True:
password1 = input('Please enter a password: ')
password2 = input('Please re-enter your password: ')
if password1 == password2:
print('Your passwords match.')
break
else:
print('Sorry. Your passwords did not match. Please try again')
user_names = []
user_names.append(userName)
with open('usernameList.txt', 'wb') as f_username:
pickle.dump(user_names, f_username)
passwords = []
passwords.append(password1)
with open('passwordList.txt', 'wb') as f_password:
pickle.dump(passwords, f_password)