«

I'm doing a replace, like this:

$text = "Yes dear!";
$pattern = "(D|d)ear";
$replace = "$1eer";

$text =~ s/$pattern/$replace/gi;

That's just an example, of course; the real $pattern and $replace come from a database list, and $text comes from form data.

The problem I'm having is that the replace is replacing with a literal "$1eer", instead of setting the $1 to (D|d). Meaning, instead of printing:

Yes deer!

I'm printing:

Yes $1eer!

Any suggestions on how to make $1 in $replace refer to the first group in $pattern?

Jason C <> writes:

> $text = "Yes dear!"; $pattern = "(D|d)ear"; $replace = "$1eer";
>
> $text =~ s/$pattern/$replace/gi;

Using this code I get "Yes eer!" in $text...

> Any suggestions on how to make $1 in $replace refer to the first group
> in $pattern?

You need to look at the /e modifier to your substitution.

//Makholm

On Monday, September 10, 2012 3:57:47 AM UTC-4, Peter Makholm wrote:
> > $text = "Yes dear!"; $pattern = "(D|d)ear"; $replace = "$1eer";
> > $text =~ s/$pattern/$replace/gi;
>
> Using this code I get "Yes eer!" in $text...

Could be a minor variation in what I posted vs. my actual code. I didn't post the whole thing because I thought it was unnecessarily complicated, but it's technically:

my $sth = $dbh->prepare("SELECT * FROM table");
$sth->execute();

while (($pattern, $replace) = $sth->fetchrow_array()) {
$text =~ s/(\b*)$pattern(er|in|ing|s|ed|y|\b)/$1$replace$+/gi;
}


> > Any suggestions on how to make $1 in $replace refer to the first group
> > in $pattern?
>
> You need to look at the /e modifier to your substitution.

Thanks for the tip. I've read a bit on the 'e' modifier now, but I'm not quite understanding how to use it for this application.

In retrospect, what I think is happening is that the while() loop is treating $replace as if it is in a single quote instead of double. So instead of it reading like:

$pattern = "(D|d)ear";
$replace = "$1eer";

it's reading:

$pattern = '(D|d)ear';
$replace = '$1eer';

So the question may really be, how do I get it to read $replace as interpretive?

Am 10.09.2012 11:36, schrieb Jason C:
>>> Any suggestions on how to make $1 in $replace refer to the first group
>>> in $pattern?
>>
>> You need to look at the /e modifier to your substitution.
>
> Thanks for the tip. I've read a bit on the 'e' modifier now, but I'm not quite understanding how to use it for this application.

For example like this:

$ perl -E '$r=q("${1}eer");($_="hello")=~s/(ll)/$r/ee; say'
helleero

- Wolf

Wolf Behrenhoff <> writes:

> For example like this:
>
> $ perl -E '$r=q("${1}eer");($_="hello")=~s/(ll)/$r/ee; say'
> helleero

So, after matching 'll' and asigning it to $1 it is replaced by

eval( eval '$r' )

Start by computing the inner eval we get

eval ( '"$1eer"')

Remembering that $1 was "ll" this evaluates to

"lleer"

//Makholm

On Monday, September 10, 2012 2:36:22 AM UTC-7, Jason C wrote:
> On Monday, September 10, 2012 3:57:47 AM UTC-4, Peter Makholm wrote:
>
> > > $text = "Yes dear!"; $pattern = "(D|d)ear"; $replace = "$1eer";
>
> > > $text =~ s/$pattern/$replace/gi;
>
> >
>
> > Using this code I get "Yes eer!" in $text...
>
>
>
> Could be a minor variation in what I posted vs. my actual code. I didn't post the whole thing because I thought it was unnecessarily complicated, but it's technically:
>
>
>
> my $sth = $dbh->prepare("SELECT * FROM table");
>
> $sth->execute();
>
>
>
> while (($pattern, $replace) = $sth->fetchrow_array()) {
>
> $text =~ s/(\b*)$pattern(er|in|ing|s|ed|y|\b)/$1$replace$+/gi;
>
> }
>
>
>
>
>
> > > Any suggestions on how to make $1 in $replace refer to the first group
>
> > > in $pattern?
>
> >
>
> > You need to look at the /e modifier to your substitution.
>
>
>
> Thanks for the tip. I've read a bit on the 'e' modifier now, but I'm not quite understanding how to use it for this application.
>
>
>
> In retrospect, what I think is happening is that the while() loop is treating $replace as if it is in a single quote instead of double. So instead of it reading like:
>
>
>
> $pattern = "(D|d)ear";
>
> $replace = "$1eer";
>
>
>
> it's reading:
>
>
>
> $pattern = '(D|d)ear';
>
> $replace = '$1eer';
>
>
>
> So the question may really be, how do I get it to read $replace as interpretive?

One way to avoid an 'ee' solution's drawbacks
is just pull the backref out of the pattern:

my $pattern = '(D|d)ear';
my $replace = 'eer';

$text =~ s/$pattern/$1$replace/gi;

--
Charles DeRykus

On Tuesday, September 11, 2012 4:50:20 PM UTC-4, C.DeRykus wrote:

> One way to avoid an 'ee' solution's drawbacks
> is just pull the backref out of the pattern:
>
> my $pattern = '(D|d)ear';
> my $replace = 'eer';
>
> $text =~ s/$pattern/$1$replace/gi;

That was my original thought, too, but I also have rows where the () isn't at the beginning. Eg:

$pattern = 'smart(\s)*ass';
$replace = 'smart$1butt';

I really would like to avoid using /ee, though, for the security reasons mentioned earlier.

Maybe something like:

$text = "Yes dear!";
$pattern = '(D|d)ear';
$replace = '$1eer';

# if $pattern doesn't contain a backreference
# create an empty one
if ($pattern !~ /\(.*?\)/g) {
$pattern = "()*?" . $pattern;
}

$replace =~ s/\$1/<marker>/g;
# now, $replace = '<marker>eer';

while ($text =~ /$pattern/g) {
$replace =~ s/<marker>/$1/g;
$text =~ s/$pattern/$replace/gi;
}


I haven't tested that, I'm just spit-balling the logic. Thoughts?

On Tuesday, September 11, 2012 7:55:15 PM UTC-7, Jason C wrote:
> On Tuesday, September 11, 2012 4:50:20 PM UTC-4, C.DeRykus wrote:
>
>
>
> > One way to avoid an 'ee' solution's drawbacks
>
> > is just pull the backref out of the pattern:
>
> >
>
> > my $pattern = '(D|d)ear';
>
> > my $replace = 'eer';
>
> >
>
> > $text =~ s/$pattern/$1$replace/gi;
>
>
>
> That was my original thought, too, but I also have rows where the () isn't at the beginning. Eg:
>
>
>
> $pattern = 'smart(\s)*ass';
>
> $replace = 'smart$1butt';
>
>
>
> I really would like to avoid using /ee, though, for the security reasons mentioned earlier.
>
>
>
> Maybe something like:
>
>
>
> $text = "Yes dear!";
>
> $pattern = '(D|d)ear';
>
> $replace = '$1eer';
>
>
>
> # if $pattern doesn't contain a backreference
>
> # create an empty one
>
> if ($pattern !~ /\(.*?\)/g) {
>
> $pattern = "()*?" . $pattern;
>
> }
>
>
>
> $replace =~ s/\$1/<marker>/g;
>
> # now, $replace = '<marker>eer';
>
>
>
> while ($text =~ /$pattern/g) {
>
> $replace =~ s/<marker>/$1/g;
>
> $text =~ s/$pattern/$replace/gi;
>
> }
>
>
>
>
>
> I haven't tested that, I'm just spit-balling the logic. Thoughts?

I'm not sure I follow entirely but, IMO, separate regexes would be much easier and more maintainable
than trying to do this in a single regex.

Only if there's a huge bottleneck, would I bother,
trying to re-factor...

--
Charles DeRykus

On Wednesday, September 12, 2012 12:08:32 AM UTC-4, C.DeRykus wrote:
> I'm not sure I follow entirely but, IMO, separate regexes would be much easier and more maintainable
>
> than trying to do this in a single regex.
>
> Only if there's a huge bottleneck, would I bother,
> trying to re-factor...

You might have missed it before, but on the live site, $pattern and $replace are coming from a database. Like so:

my $sth = $dbh->prepare("SELECT * FROM table");
$sth->execute();

while (($pattern, $replace) = $sth->fetchrow_array()) {
$text =~ s/(\b*)$pattern(er|in|ing|s|ed|y|\b)/$1$replace$+/gi;
}

The first group in $pattern can actually be anywhere in the string, so one row might be:

(D|d)ear

while the next might be:

smart(\s*)ass

The issue comes in where $1 is defined as non-interpretive in the database, and I'm not sure how to make it interpretive in the replacement.

The while() loop that I presented in the last post is an attempt to replace the non-interpretive '$1' with '<marker>', then replace '<marker>' back with the interpretive "$1".

Jason C wrote:
) On Tuesday, September 11, 2012 4:50:20 PM UTC-4, C.DeRykus wrote:
)
)> One way to avoid an 'ee' solution's drawbacks
)> is just pull the backref out of the pattern:
)>
)> my $pattern = '(D|d)ear';
)> my $replace = 'eer';
)>
)> $text =~ s/$pattern/$1$replace/gi;
)
) That was my original thought, too, but I also have rows where the () isn't at the beginning. Eg:
)
) $pattern = 'smart(\s)*ass';
) $replace = 'smart$1butt';
)
) I really would like to avoid using /ee, though, for the security reasons mentioned earlier.
)
) Maybe something like:
)
) $text = "Yes dear!";
) $pattern = '(D|d)ear';
) $replace = '$1eer';
)
) # if $pattern doesn't contain a backreference
) # create an empty one
) if ($pattern !~ /\(.*?\)/g) {
) $pattern = "()*?" . $pattern;
) }
)
) $replace =~ s/\$1/<marker>/g;
) # now, $replace = '<marker>eer';
)
) while ($text =~ /$pattern/g) {
) $replace =~ s/<marker>/$1/g;
) $text =~ s/$pattern/$replace/gi;
) }

It would be easier to do the whole thing in a /e expression.
But not interpreting the database string, but just adding your own code.

Like this:

$test =~ s/$pattern/my $s1 = $1; (my $t = $replace) =~ s|\$1|$s1|g; $t/ge;

That should work.

If you want more than just $1, you need a slightly more complicated
expression, probably involving @- and @+.

(I've always wondered why there is no regex-match array perlvar...)


SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT