regular expression to represent time in a special format

I have many files containing date extensions. For example, today's file would be named

filename.20101118

and format is yyyymmdd. I开发者_如何学编程 would like to list/grep/etc for all dated files.

Do you mean just something like this ?

/filename\.[12][90][0-9][0-9][01][0-9][0-3][0-9]/

Note that this matches non-valid dates, too.

or, more generalized:

/[a-zA-Z0-9_]+\.[12][90][0-9][0-9][01][0-9][0-3][0-9]/

instead of [a-zA-Z0-9_] you can use whatever characters your filename can contain.

Try this if you're not looking to do any validation.

\.(\d{8})

Note: I have assumed that your "filename" will be only alphanumeric (ASCII).

My best friend RegexBuddy says:

[a-zA-Z0-9]+?\.(19|20)[0-9]{2}(0[1-9]|1[012])(0[1-9]|[12][0-9]|3[01])

Matches

filename123.20101118
FOOBAR123.19961212

Does not match

FOOBAR.88881201
foobar.20103512
filename.20101235

Disclaimer: I am just a happy user and have no affiliation to RB

Explanation

[a-zA-Z0-9]+?\.(19|20)[0-9]{2}(0[1-9]|1[012])(0[1-9]|[12][0-9]|3[01])

Options: case insensitive

Match a single character present in the list below «[a-zA-Z0-9]+?»
   Between one and unlimited times, as few times as possible, expanding as needed (lazy) «+?»
   A character in the range between “a” and “z” «a-z»
   A character in the range between “A” and “Z” «A-Z»
   A character in the range between “0” and “9” «0-9»
Match the character “.” literally «\.»
Match the regular expression below and capture its match into backreference number 1 «(19|20)»
   Match either the regular expression below (attempting the next alternative only if this one fails) «19»
      Match the characters “19” literally «19»
   Or match regular expression number 2 below (the entire group fails if this one fails to match) «20»
      Match the characters “20” literally «20»
Match a single character in the range between “0” and “9” «[0-9]{2}»
   Exactly 2 times «{2}»
Match the regular expression below and capture its match into backreference number 2 «(0[1-9]|1[012])»
   Match either the regular expression below (attempting the next alternative only if this one fails) «0[1-9]»
      Match the character “0” literally «0»
      Match a single character in the range between “1” and “9” «[1-9]»
   Or match regular expression number 2 below (the entire group fails if this one fails to match) «1[012]»
      Match the character “1” literally «1»
      Match a single character present in the list “012” «[012]»
Match the regular expression below and capture its match into backreference number 3 «(0[1-9]|[12][0-9]|3[01])»
   Match either the regular expression below (attempting the next alternative only if this one fails) «0[1-9]»
      Match the character “0” literally «0»
      Match a single character in the range between “1” and “9” «[1-9]»
   Or match regular expression number 2 below (attempting the next alternative only if this one fails) «[12][0-9]»
      Match a single character present in the list “12” «[12]»
      Match a single character in the range between “0” and “9” «[0-9]»
   Or match regular expression number 3 below (the entire group fails if this one fails to match) «3[01]»
      Match the character “3” literally «3»
      Match a single character present in the list “01” «[01]»


Created with RegexBuddy

Simplest:

find -regex ".*/.*\.[0-9]+$"

Better:

find -regextype posix-basic -regex ".*/.*\.[0-9]\{8\}$"

Best?:

find -regextype posix-extended -regex ".*/.*\.[0-9]{4}([0][1-9]|1[012])(0[1-9]|[12][0-9]|3[01])$"

The last one will find any year (we don't want a repeat of Y2K, now, do we?). and only months in the range 01-12 and days in the range 01-31. Of course, it doesn't validate for the correct number of days in a month or for leap days.