Perl: Matching string not containing PATTERN

Question

While using Perl regex to chop a string down into usable pieces I had the need to match everything except a certain pattern. I solved it after I found this hint on Perl Monk

Answer 1

Building it up piece by piece (and throughout assuming no newlines in the string or PATTERN):

This matches any string:

/^.*$/

But we don't want . to match a character that starts PATTERN, so replace

.

with

(?!PATTERN).

This uses a negative look-ahead that tests a given pattern without actually consuming any of the string and only succeeds if the pattern does not match at the given point in the string. So it's like saying:

if PATTERN doesn't match at this point,
    match the next character

This needs to be done for every character in the string, so * is used to match zero or more times, from the beginning to the end of the string.

To make the * apply to the combination of the negative look-ahead and ., not just the ., it needs to be surrounded by parentheses, and since there's no reason to capture, they should be non-capturing parentheses (?: ):

(?:(?!PATTERN).)*

And putting back the anchors to make sure we test at every position in the string:

/^(?:(?!PATTERN).)*$/

Note that this solution is particularly useful as part of a larger match; e.g. to match any string with foo and later baz but no bar in between:

/foo(?:(?!bar).)*baz/

If there aren't such considerations, you can simply do:

/^(?!.*PATTERN)/

to check that PATTERN does not match anywhere in the string.

About newlines: there are two problems with your regex and newlines. First, . doesn't match newlines, so "foo\nbar" =~ /^(?:(?!baz).)*$/ doesn't match, even though the string does not contain baz. You need to add the /s flag to make . match any character; "foo\nbar" =~ /^(?:(?!baz).)*$/s correctly matches. Second, $ doesn't match just at the end of the string, it also can match before a newline at the end of the string. So "foo\n" =~ /^(?:(?!\s).)*$/s does match, even though the string contains whitespace and you are attempting to only match strings with no whitespace; \z always only matches at the end, so "foo\n" =~ /^(?:(?!\s).)*\z/s correctly fails to match the string that does in fact contain a \s. So the correct general purpose regex is /^(?:(?!PATTERN).)*\z/s