responses embedded
[EMAIL PROTECTED] wrote on 07/19/2006 06:25:50
PM:
> hi josh --
>
> In a message dated 7/19/2006 4:22:20 P.M. Eastern Standard Time,
> [EMAIL PROTECTED] writes:
>
> > Gurus-
> >
> > I'm trying to help a co-worker on reg exp/pattern match issue.
> >
> > we have a set of items to go through and another set of input that's
> > tested against the first.
> >
> > we need to check that items in the second set are a one-for-one match
or
> > subset of the latter.
> >
> >
> > ie:
> >
> > $var1 =~ /$var2/
> >
> > where this returns true in a case such as $var1=Jose, Joseph, and Josh
and
> > $var2 will be James and Joseph
> >
> > when $var2 is James they need to return false and when it is Joseph
both
> > Jose and Joseph should return true while Josh returns false.
> >
> > So far we haven't found a solution in orielly and trying
> >
> > if( $var1 =~ m/$var2/) { print "$var2 contains $var1\n"; }
> > else { print "failure on $var2 and $var1\n"; }
> >
> >
> > has failed to work.
> >
> > -Josh
>
> i may not understand this correctly, it seems to me that you are
> saying that items
> in the first set (represented by $var1) will always be equal to or
> less than the length
> of items in the second set (represented by $var2). i'm also
> assuming that by
> ``subset'' you mean ``subset anchored at the beginning of the string''.
>
> if that is so, it seems that the way to go about comparing the two
> sets is the opposite
> of the example you give, e.g.,
> $var2 =~ /$var1/;
> since we don't really care if part of $var2 slops over on the end.
>
> maybe something like this:
>
> C:[EMAIL PROTECTED]>perl -we "use strict; my $v1 = shift; for my $v2
(@ARGV) {
> my $re = qr/$v2/x;
> print qq($v1 ), ($v1 =~ / ^ $re /x) ? q(contains) : q(does NOT
> contain), qq( $v2 \n) }"
> Joseph Jose Joseph Josh Josephhh
> Joseph contains Jose
> Joseph contains Joseph
> Joseph does NOT contain Josh
> Joseph does NOT contain Josephhh
>
> note the ^ beginning of string anchor; easy enough to take this out
> and have the match
> be anywhere in the string.
Yeah, the example fails in the sub string may not start at the beginning.
How it's being attained is also a bit more complex. Than you for your
answer. The hints you, the other Bill, et al, have given so far have
gotten us playing around and have found some methods that partially work.
We're going to try to manipulate those to something that does. At that
point, If Joe feels okay with it I'll ask he post the solution we found.
If we decide we need more help I'll see if we can post part of the code to
show how the two variables are attained. Maybe that could help too when a
solution is found. maybe someone might even have a far more elegant and
simple solution than we find.
> there's also a version that doesn't involve regexes:
>
> C:[EMAIL PROTECTED]>perl -we "use strict; my $v1 = shift; for my $v2
(@ARGV) {
> print qq($v1 ), (index($v1, $v2) >= 0) ? q(contains) : q(does NOT
> contain), qq( $v2 \n) }"
> Joseph Jose Joseph Josh Josephhh
> Joseph contains Jose
> Joseph contains Joseph
> Joseph does NOT contain Josh
> Joseph does NOT contain Josephhh
>
> hth -- bill walters
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