Re^2: Quick Regex Question

[Anonymous Monk]

Thank you for the quick reply. I must be doing something wrong as now it matches 7,8,9,10, etc numbers. I am looking to only match exactly 7?

[Athanasius]

OK, change the regex to:

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

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)/

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

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

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

[GrandFather]

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.

[AnomalousMonk]

... using a zero-width negative look-ahead assertion; or ... using a z
+ero-width positive look-ahead assertion together with \D ...

Note that  (?!\d) and  (?=\D) are not complementaryequivalent. 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

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.