如何用正则验证MMDDYY格式无分隔符日期？含闰年场景

Got it, let's work through this regex problem—since you already know you can use DateTime.ParseExact but specifically want a regex solution, we need to fix the gaps in your original pattern: it didn't account for month-specific day limits (like April only having 30 days) or leap year exceptions for February 29th.

The Problem with Your Original Regex #

Your original pattern ^(|(0[1-9])|(1[0-2]))((0[1-9])|(1\d)|(2\d)|(3[0-1]))((\d{2}))$ just checks that the month is 01-12 and day is 01-31, which fails for edge cases like 023020 (invalid, February has no 30th) or 022919 (invalid, 2019 isn't a leap year).

The Solution: A Regex That Handles All Cases #

Here's a regex that properly validates MMDDYY by breaking down month-day rules and leap year logic for February:

^(?:
  (0[13578]|1[02])(0[1-9]|[12]\d|3[01])\d{2}  # Months with 31 days (Jan,Mar,May,Jul,Aug,Oct,Dec)
  |
  (0[469]|11)(0[1-9]|[12]\d|30)\d{2}          # Months with 30 days (Apr,Jun,Sep,Nov)
  |
  02(0[1-9]|[12]\d|2[0-8])\d{2}               # February (non-leap year: 01-28)
  |
  0229(00|([02468][048]|[13579][26]))         # February 29 (only leap years: divisible by 4)
)$

How It Works #

Let's break down each branch:

• 31-day months: Matches months 01,03,05,07,08,10,12 followed by days 01-31, then any two-digit year.
• 30-day months: Matches months 04,06,09,11 followed by days 01-30, then any two-digit year.
• Non-leap February: Matches 02 followed by days 01-28, then any two-digit year.
• Leap February: Matches 0229 only if the two-digit year is divisible by 4. The pattern (00|([02468][048]|[13579][26])) covers all two-digit years divisible by 4 (e.g., 20 for 2020, 00 for 2000, 96 for 1996).

Notes #

• Leap Year Edge Case: This regex assumes two-digit years like 00 correspond to 2000 (a leap year). If your use case treats 00 as 1900 (a non-leap year), you'd need to adjust the leap year branch to exclude 00—but this is rare in most modern applications.
• Compressed Version: If you need a one-line regex without comments, use this:

  ^(?:(0[13578]|1[02])(0[1-9]|[12]\d|3[01])\d{2}|(0[469]|11)(0[1-9]|[12]\d|30)\d{2}|02(0[1-9]|[12]\d|2[0-8])\d{2}|0229(00|([02468][048]|[13579][26])))$
  

Test Cases #

• Valid: 112517, 022920, 043099, 123100
• Invalid: 023020, 022919, 130100, 063121

内容的提问来源于stack exchange，提问作者sruti