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Re: Re: Matching brackets in Regular Expression

by jweed (Chaplain)
on Nov 21, 2003 at 18:28 UTC ( [id://308996]=note: print w/replies, xml ) Need Help??


in reply to Re: Matching brackets in Regular Expression
in thread Matching brackets in Regular Expression

The problem is that this scalar $stub is volatile and it also contains a little bracket...

I believe that though he listed that bob(scalar), it should be bob[scalar] as the description lists. In this case it would be interpreted as a character class instead of a string. \Q and \E would backslash these brackets.

In it's current form, there's really no conflict unless the string is something like bob(??{print $hi}).



Who is Kayser Söze?
All code is untested unless I say so.

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Re: Re: Re: Matching brackets in Regular Expression
by qq (Hermit) on Nov 21, 2003 at 22:54 UTC

    Except that the parens () will be used for grouping and capturing by the regex, and so won't match the literals in the string; 

    perl -e '$s = q|b(s)|; print  "without quotemeta" if $s =~ /$s/; print
+ "with quotemeta" if $s =~ /\Q$s\E/;'

with quotemeta

    [download]

    qq

[reply]
[d/l]

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