01密码强度检查_python篇

需求：

 

 

 

解决方法（本人写的）：

写一个与需求相关的正则表达式，断言是否匹配，不匹配返回false，匹配返回ture

 

代码参考如下

 

import re

def checkio(data):
    password_one = len(data) >= 10  # 长度大于等于10
    password_two = re.findall("[0-9]",data) # 必有一个数字
    password_three = re.findall("[a-z]",data)   # 必有一个小写字母
    password_four = re.findall("[A-Z]",data)    # 必有一个大写字母
    if password_one and password_two and password_three and password_four:
        return True
    else:
        return False

# 断言
if __name__ == '__main__':
    assert checkio('A1213pokl') == False, "1st example"
    assert checkio('bAse730onE4') == True, "2nd example"
    assert checkio('asasasasasasasaas') == False, "3rd example"
    assert checkio('QWERTYqwerty') == False, "4th example"
    assert checkio('123456123456') == False, "5th example"
    assert checkio('QwErTy911poqqqq') == True, "6th example"
    assert  checkio("erer798rew9rew9r7ew987rw") == False

# 成功的话，就输出这句话
    print("Coding complete? Click 'Check' to review your tests and earn cool rewards!")

 

 

 

其他解决方法：

参考他人的，这个正则表达式没看懂，有理解的同学麻烦留言一下

import re

 

 

def checkio(data):

    # (?=) is positive lookahead. Use this construction to make sure the enclosed pattern exists

    return bool(re.search(r'^(?=.*\d)(?=.*[a-z])(?=.*[A-Z]).{10,}$', data))

 

if __name__ == '__main__':

    # These "asserts" using only for self-checking and not necessary for

    # auto-testing

    assert checkio('A1213pokl') == False, "1st example"

    assert checkio('bAse730onE4') == True, "2nd example"

    assert checkio('asasasasasasasaas') == False, "3rd example"

    assert checkio('QWERTYqwerty') == False, "4th example"

    assert checkio('123456123456') == False, "5th example"

    assert checkio('QwErTy911poqqqq') == True, "6th example"