Regex to find 3 out of 4 conditions

问题:

My client has requested that passwords on their system must following a specific set of validation rules, and I'm having great difficulty coming up with a "nice" regular expression.

The rules I have been given are...

• Minimum of 8 character
• Allow any character
• Must have at least one instance from three of the four following character types...
  1. Upper case character
  2. Lower case character
  3. Numeric digit
  4. "Special Character"

When I pressed more, "Special Characters" are literally everything else (including spaces).

I can easily check for at least one instance for all four, using the following...

^(?=.*?[A-Z])(?=.*?[a-z])(?=.*?\d)(?=.*?[^a-zA-Z0-9]).{8,}$

The following works, but it's horrible and messy...

^((?=.*?[A-Z])(?=.*?[a-z])(?=.*?\d)|(?=.*?[A-Z])(?=.*?[a-z])(?=.*?[^a-zA-Z0-9])|(?=.*?[A-Z])(?=.*?\d)(?=.*?[^a-zA-Z0-9])|(?=.*?[a-z])(?=.*?\d)(?=.*?[^a-zA-Z0-9])).{8,}$

So you don't have to work it out yourself, the above is checking for (1,2,3|1,2,4|1,3,4|2,3,4) which are the 4 possible combinations of the 4 groups (where the number relates to the "types" in the set of rules).

Is there a "nicer", cleaner or easier way of doing this?

(Please note, this is going to be used in an <asp:RegularExpressionValidator> control in an ASP.NET website, so therefore needs to be a valid regex for both .NET and javascript.)

回答1:

It's not much of a better solution, but you can reduce [^a-zA-Z0-9] to [\W_], since a word character is all letters, digits and the underscore character. I don't think you can avoid the alternation when trying to do this in a single regex. I think you have pretty much have the best solution.

One slight optimization is that \d*[a-z]\w_*|\d*[A-Z]\w_* ~> \d*[a-zA-Z]\w_*, so I could remove one of the alternation sets. If you only allowed 3 out of 4 this wouldn't work, but since \d*[A-Z][a-z]\w_* was implicitly allowed it works.

(?=.{8,})((?=.*\d)(?=.*[a-z])(?=.*[A-Z])|(?=.*\d)(?=.*[a-zA-Z])(?=.*[\W_])|(?=.*[a-z])(?=.*[A-Z])(?=.*[\W_])).*

Extended version:

(?=.{8,})(
  (?=.*\d)(?=.*[a-z])(?=.*[A-Z])|
  (?=.*\d)(?=.*[a-zA-Z])(?=.*[\W_])|
  (?=.*[a-z])(?=.*[A-Z])(?=.*[\W_])
).*

REY

回答2:

I'd like to improve the accepted solution with this one

^(?=.{8,})(
    (?=.*[^a-zA-Z\s])(?=.*[a-z])(?=.*[A-Z])|
    (?=.*[^a-zA-Z0-9\s])(?=.*\d)(?=.*[a-zA-Z])
).*$

回答3:

The above Regex worked well for most scenarios except for strings such as "AAAAAA1$", "$$$$$$1a" This could be an issue only in iOS ( Objective C and Swift) that the regex "\d" has issues The following fix worked in iOS, i.e changing to [0-9] for digits

^((?=.*?[A-Z])(?=.*?[a-z])(?=.*?[0-9])|(?=.*?[A-Z])(?=.*?[a-z])(?=.*?[^a-zA-Z0-9])|(?=.*?[A-Z])(?=.*?[0-9])(?=.*?[^a-zA-Z0-9])|(?=.*?[a-z])(?=.*?[0-9])(?=.*?[^a-zA-Z0-9])).{8,}$

回答4:

Password must meet at least 3 out of the following 4 complexity rules,

[at least 1 uppercase character (A-Z) at least 1 lowercase character (a-z) at least 1 digit (0-9) at least 1 special character — do not forget to treat space as special characters too]

at least 10 characters

at most 128 characters

not more than 2 identical characters in a row (e.g., 111 not allowed)

'^(?!.(.)\1{2}) ((?=.[a-z])(?=.[A-Z])(?=.[0-9])|(?=.[a-z])(?=.[A-Z])(?=.[^a-zA-Z0-9])|(?=.[A-Z])(?=.[0-9])(?=.[^a-zA-Z0-9])|(?=.[a-z])(?=.[0-9])(?=.*[^a-zA-Z0-9])).{10,127}$'

(?!.*(.)\1{2})

(?=.[a-z])(?=.[A-Z])(?=.*[0-9])

(?=.[a-z])(?=.[A-Z])(?=.*[^a-zA-Z0-9])

(?=.[A-Z])(?=.[0-9])(?=.*[^a-zA-Z0-9])

(?=.[a-z])(?=.[0-9])(?=.*[^a-zA-Z0-9])

.{10,127}