python - 输入没有选择密码长度

Question

我是编程新手，我正在做一个密码管理器。我的问题：-它的作用->当我运行程序并输入密码的长度时，它会生成一个密码，但它始终是 4 位数字，并且密码重复的次数与我输入的数字一样多。-它应该做什么 -> 我输入的数字应该决定密码的长度，而不是重复多少次。

import random

#shuffle the list
def shuffle(string):
  tempList = list(string)
  random.shuffle(tempList)
  return ''.join(tempList)

#the password functions
uppercaseLetter=chr(random.randint(65,90))
lowercaseLetter=chr(random.randint(97,122))
punctuationSign=chr(random.randint(32,152))
digit=chr(random.randint(48,57))

#completing the password
passwordLength = int(input("choose your password length: "))

possibleChars = uppercaseLetter, lowercaseLetter, punctuationSign, digit
ranChar = shuffle(possibleChars)

tempPassword = []

count = 0
while count != passwordLength:
    tempPassword.extend(ranChar)
    count = count + 1

password = ''.join(tempPassword)

#for which sitename
sitename = input('Save under which name: ')


print (password, sitename)

data=open("test.txt",'a')
data.write(sitename +'  ')
data.write(password +'\n')
data.close()

Answer 1

我认为你真的把你需要做的事情复杂化了，据我所知，这是你想要做的事情

import random

#completing the password
passwordLength = int(input("choose your password length: "))

temp_pass = ''

count = 0
while count != passwordLength:
    char_type = random.randint(0,3)
    if char_type == 0:
        random_char = chr(random.randint(65,90))
    elif char_type == 1:
        random_char = chr(random.randint(97,122))
    elif char_type == 2:
        random_char = chr(random.randint(32,152))
    else:
        random_char = chr(random.randint(48,57))
    temp_pass = temp_pass + random_char
    count = count + 1


print(temp_pass)

希望这可以帮助。我建议在尝试这样的事情之前多练习基础知识，代码中有很多不好的做法。