用正则表达式匹配“神奇”的日期

Question

I've found a regular expression which matches "magical" dates (in which the last two digits of the year are the same as the two digits of the month and day, for example 2008-08-08):

\b[0-9][0-9]\([0-9][0-9])-\1-\1\b

... but I can't understand it. How does it work?

Answer 1

Here's the same regex, written verbosely with comments:

\b            # The beginning or end of a word.
[0-9]         # Any one of the characters '0'-'9'.
[0-9]         # Any one of the characters '0'-'9'.
(             # Save everything from here to the matching ')' in a variable '\1'.
    [0-9]     # Any one of the characters '0'-'9'.
    [0-9]     # Any one of the characters '0'-'9'.
)             # 
-             # The literal character '-'
\1            # Whatever was saved earlier, between the parentheses.
-             # The literal character '-'
\1            # Whatever was saved earlier, between the parentheses.
\b            # The beginning or end of a word.

In the case of '2008-08-08', the '20' gets matched by the first two [0-9] s, and then the '08' immediately after that gets matched by the next two [0-9] s (which are in parentheses, so that '08' gets saved to the variable \\1 ).

Then a hypen is matched, then 08 again (because it was stored in the variable \\1 earlier), then another hyphen, then 08 (as \\1 ) again.

Answer 2

You can use regex:

\b[0-9]{2}([0-9]{2})-\1-\1\b

Last 2 digits are captured in captured group #1 and then back-reference of captured group ie \\1 is used in month and date part later.

RegEx Demo