Thomas Hartman wrote:
testPcre = ( subRegex (mkRegex "(?<!\n)\n(?!\n)") "asdf\n \n\n\nadsf"
"" ) == "asdf \n\n\nadsf"
quoting from the man page for regcomp:
REG_NEWLINE Compile for newline-sensitive matching. By default, newline is a
completely ordinary character with
no special meaning in either REs or strings. With this flag,
`[^' bracket expressions and `.' never
match newline, a `^' anchor matches the null string after any
newline in the string in addition to
its normal function, and the `$' anchor matches the null string
before any newline in the string in
addition to its normal function.
This is the carried over to Text.Regex with
mkRegexWithOpts Source
:: String The regular expression to compile
-> Bool True <=> '^' and '$' match the beginning and end of individual
lines respectively, and '.' does not match the newline character.
-> Bool True <=> matching is case-sensitive
-> Regex Returns: the compiled regular expression
Makes a regular expression, where the multi-line and case-sensitive options can
be changed from the default settings.
Or with regex-posix directly the flag is "compNewline":
http://hackage.haskell.org/packages/archive/regex-posix/0.94.1/doc/html/Text-Regex-Posix-Wrap.html
> The defaultCompOpt is (compExtended .|. compNewline).
You want to match a \n that is not next to any other \n.
So you want to turn off REG_NEWLINE.
import Text.Regex.Compat
r :: Regex
r = mkRegexWithOpts "(^|[^\n])\n($|[^\n])" False True -- False is important
here
The ^ and $ take care of matching a lone newline at the start or end of the
whole text. In the middle of the text the pattern is equivalent to [^\n]\n[^\n].
When substituting you can use the \1 and \2 captures to restore the matched
non-newline character if one was present.
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