Python 数字、字母出现次数的正则表达式

Question

我正在尝试编写正则表达式来验证令牌。

my token has a following structure:

3个或更多[0-9]digits
5 个或更多 [a-z] small alpha
2个或更多[#@] special characters
5 个或更多 [A-Z] upper alpha（可选 ?）可能存在或不存在。

注意：可选的 [A-Z] 字符可能存在于某些标记中，也可能不存在，当它存在于标记中时，它应该在 5 or more 次之间。

令牌长度不同于 minimum of 6 to a max of 30

few examples of valid/invalid matches:

token1 : t4xa@ui13p#o6
breakdown : there are 7 [a-z] , 2 special chracter[@#] , 4 digits [0-9]
VALID MATCH: True

token2: 3@piy13Qx9#13@z1337
breakdown: there are 5 [a-z] , 3 special character [@#] , 10 digits [0-9] and 1 [A-Z](which is optional)
VALID MATCH: False (because `[A-Z]` exist in token but it exist for `1` time, it should exist atleast of 5 or more.)

token3: 3@piy1ABC3Qx9#13@DEGFz1337
breakdown: there are 5[a-z], 7 [A-Z] , 10 digits [0-9]
VALID MATCH: True

到目前为止，我设法编写了一个正则表达式来验证至少一位数字、一个大写字母、至少一个小写字母、至少一个特殊字符。

^(?=.*[\d])(?=.*[A-Z])(?=.*[a-z])(?=.*[@#])[\w\d@#]{6,30}$

Answer 1

如果还有 0 个或 5 个大写字符，那么您可以使用正向先行断言没有大写字符或至少 5 次。

(?=(?:[^A-Z\s]*$|(?:[^A-Z\s]*[A-Z]){5})

您可以使用否定字符类和量词来检查最少出现次数。

^(?=(?:[^\d\s]*\d){3})(?=(?:[^a-z\s]*[a-z]){5})(?=(?:[^@#\s]*[@#]){2})(?=(?:[^A-Z\s]*$|(?:[^A-Z\s]*[A-Z]){5}))[\w\d@#]{6,30}$

Regex demo

Python 数字、字母出现次数的正则表达式

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