Re^3: Quick Regex Question

OK, change the regex to:

/INC\d{7}(?:[^\d]|$)/
[download]

which should match only exactly 7 digits. (INC followed by 7 digits followed by either a non-digit or the end of the string.)

Update: Better solutions:

/INC\d{7}(?!\d)/
[download]

using a zero-width negative look-ahead assertion; or:

/INC\d{7}(?:(?=\D)|$)/    # Corrected: See post by AnomalousMonk, belo
+w.
[download]

using a zero-width positive look-ahead assertion together with \D as per GrandFather’s suggestion. See Extended Patterns.

Athanasius <°(((>< contra mundum

Comment on Re^3: Quick Regex Question Select or Download Code
Re^4: Quick Regex Question by GrandFather (Cardinal) on Oct 08, 2012 at 23:29 UTC
You can use \D for "not a digit". In like fashion you can use \W and \S to complement the common word and space matches. True laziness is hard work	[reply]
Re^4: Quick Regex Question by AnomalousMonk (Prior) on Oct 09, 2012 at 12:36 UTC
`... using a zero-width negative look-ahead assertion; or ... using a z +ero-width positive look-ahead assertion together with \D ...` [download] Note that `(?!\d)` and `(?=\D)` are not ~~complementary~~equivalent. The `(?!\d)` assertion is satisfied by having any non-digit follow it or by nothing, i.e., by the end of the string, nothing being not-a-digit. The `(?=\D)` assertion must be followed by a non-digit character. And similarly with the `(?<!\d)` and `(?<=\D)` look-behinds. `>perl -wMstrict -le "my $s = 'xyzINC1234567'; ;; print 'negative assertion true' if $s =~ /INC\d{7}(?!\d)/; print 'positive assertion true' if $s =~ /INC\d{7}(?=\D)/; " negative assertion true` [download] In this behavior, `(?!\d)` and, in general, `(?!`character-or-character-class`)` and its negative look-behind cousin are similar to (but not exactly the same as) the `\b` assertion, which may be true at the beginning or end of a string.	[reply] [d/l] [select]


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