in reply to Re: Re: Re: Re: conditional match in regex
in thread conditional match in regex

Actually, $1 = type if the type is $, @, %, or *; however, in the <name> case, the only way to check its type is by nested conditionals outside the regex; something like: 

if ($2) {
    if ($1 eq '$') {
        ... 
    }
    elsif ($1 eq '@') {
        ... 
    }
    ... 
}
else {
    ... 
}

[download]

Which, IMHO, is a huge pain in the behind compared to a single if-elsif chain. Why do more work than you have to?

Replies are listed 'Best First'.
Re: Re: Re: Re: Re: Re: conditional match in regex
by petral (Curate) on Nov 06, 2002 at 14:58 UTC
    The problem seems to be stated: either $val or <val>:
    $ perl -le'$_=pop;/([\$@*%]|(<))(.*)(?(2)>)/;print"($1)($3)"' '$xyz'
($)(xyz)
$ perl -le'$_=pop;/([\$@*%]|(<))(.*)(?(2)>)/;print"($1)($3)"' '$xyz>'
($)(xyz>)
$ perl -le'$_=pop;/([\$@*%]|(<))(.*)(?(2)>)/;print"($1)($3)"' '<xyz>'
(<)(xyz)

    [download]


      p
[reply]
[d/l]
      '$xyz>' is definately NOT a valid symbol table entry.
[reply]
[d/l]
        Right.  I was just showing how the regex works.
        $ perl -le'"%xyz"=~/([\$@*%]|(<))(.*)((?(2)>))/;print"($1)($3)($4)"'
(%)(xyz)()
$ perl -le'"%xyz>"=~/([\$@*%]|(<))(.*)((?(2)>))/;print"($1)($3)($4)"'
(%)(xyz>)()
$ perl -le'"<xyz>"=~/([\$@*%]|(<))(.*)((?(2)>))/;print"($1)($3)($4)"'
(<)(xyz)(>)

        [download]

          p
[reply]
[d/l]

In Section Seekers of Perl Wisdom