Re: Re: Re: Re: Re: 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?

Comment on Re: Re: Re: Re: Re: conditional match in regex Download Code

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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)($3)"' '<xyz>' (<)(xyz)` [download] p	[reply] [d/l]
Re: Re: Re: Re: Re: Re: Re: conditional match in regex by jryan (Vicar) on Nov 06, 2002 at 16:19 UTC
`'$xyz>'` is definately NOT a valid symbol table entry.	[reply] [d/l]
Re: Re: Re: Re: Re: Re: Re: Re: conditional match in regex by petral (Curate) on Nov 06, 2002 at 16:46 UTC
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]


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