I have some code:
$a = 'abcdef';
$b = 'f\1d';
$a =~ s/d(e)f/$b/g;
print $a;
I get abcf\1d, but how can I get abcfed?
$a = 'abcdef';
$b = '"f$1d"';
$a =~ s/d(e)f/$b/gee;
print $a;
Notice that there are two e modifiers in /gee; the second one evaluates "f$1d" string to fed.
As a side note you don't need /g as you're replacing only one pattern occurrence.
$a and $b are special variables, used for sort. So you shoulnd't use them at all.
Better solution is to use another variable and then:
$var_a = 'abcdef';
$var_b = '"f${1}d"';
$var_a =~ s/d(e)f/$var_b/gee;
print $var_a;
The trick is to say, that the substitution uses the \1 first found var ( use the /e flag to avoid using \1 :P )
Related
I'm sure there is a very simple way to do this, but whenever I search for examples, I get the two step method. Here is what I typically do:
$data =~ m/(my_query)/;
$result = $1;
I want to set $result in the same line as the regex and never use $1. Thanks!
my($result) = ($data =~ m/(my_query)/);
As noted in a comment, the my($result) needs the parentheses to provide an array context for the result of the match. In an array context, you get the $1 etc allocated to the array. You could use #result = ($data =~ m/(my_query)/);; you could omit the my but you would need to keep the parentheses; you could subscript the array using $result = ($data =~ m/(my_query)/)[0]; (thanks ysth). The key words here are 'array context'.
Examples:
$ perl -e '$data="abcdef";my($result)=($data =~ m/(cde)/); print "$result\n"'
cde
$ perl -e '$data="abcdef"; ($result)=($data =~ m/(cde)/); print "$result\n"'
cde
$ perl -e '$data="abcdef"; #result =($data =~ m/(cde)/); print "$result[0]\n"'
cde
$ perl -e '$data="abcdef"; $result =($data =~ m/(cde)/)[0]; print "$result\n"'
cde
$
You didn't specify what problem you want to avoid, but there is definitely one to avoid. The following code assigns something unknown to $result when the pattern doesn't match:
$data =~ /(my_query)/;
my $result = $1;
You could use a conditional to assign something useful to $result when the pattern doesn't match
my $result = $data =~ /(my_query)/ ? $1 : undef;
Or you could take advantage of the fact that m// in list context returns what it captured.
my ($result) = $data =~ /(my_query)/;
$data="abcde";
$data =~ s/(cde)/$result=$1/e;
I need to write a perl regex to convert
site.company.com => dc=site,dc=company,dc=com
Unfortunately I am not able to remove the trailing "," using the regex I came with below. I could of course remove the trailing "," in the next statement but would prefer that to be handled as a part of the regex.
$data="site.company.com";
$data =~ s/([^.]+)\.?/dc=$1,/g;
print $data;
This above code prints:
dc=site,dc=company,dc=com,
Thanks in advance.
When handling urls it may be a good idea to use a module such as URI. However, I do not think it applies in this case.
This task is most easily solved with a split and join, I think:
my $url = "site.company.com";
my $string = join ",", # join the parts with comma
map "dc=$_", # add the dc= to each part
split /\./, $url; # split into parts
$data =~s/\./,dc=/g&&s/^/dc=/g;
tested below:
> echo "site.company.com" | perl -pe 's/\./,dc=/g&&s/^/dc=/g'
dc=site,dc=company,dc=com
Try doing this :
my $x = "site.company.com";
my #a = split /\./, $x;
map { s/^/dc=/; } #a;
print join",", #a;
just put like this,
$data="site.company.com";
$data =~ s/,dc=$1/dc=$1/g; #(or) $data =~ s/,dc/dc/g;
print $data;
I'm going to try the /ge route:
$data =~ s{^|(\.)}{
( $1 && ',' ) . 'dc='
}ge;
e = evaluate replacement as Perl code.
So, it says given the start of the string, or a dot, make the following replacement. If it captured a period, then emit a ','. Regardless of this result, insert 'dc='.
Note, that I like to use a brace style of delimiter on all my evaluated replacements.
Is there an easy way to add regex modifiers such as 'i' to a quoted regular expression? For example:
$pat = qr/F(o+)B(a+)r/;
$newpat = $pat . 'i'; # This doesn't work
The only way I can think of is to print "$pat\n" and get back (?-xism:F(o+)B(a+)r) and try to remove the 'i' in ?-xism: with a substitution
You cannot put the flag inside the result of qr that you already have, because it’s protected. Instead, use this:
$pat = qr/F(o+)B(a+)r/i;
You can modify an existing regex as if it was a string as long as you recompile it afterwards
my $pat = qr/F(o+)B(a+)r/;
print $pat, "\n";
print 'FOOBAR' =~ $pat ? "match\n" : "mismatch\n";
$pat =~ s/i//;
$pat = qr/(?i)$pat/;
print $pat, "\n";
print 'FOOBAR' =~ $pat ? "match\n" : "mismatch\n";
OUTPUT
(?-xism:F(o+)B(a+)r)
mismatch
(?-xism:(?i)(?-xsm:F(o+)B(a+)r))
match
Looks like the only way is to stringify the RE, replace (-i) with (i-) and re-quote it back:
my $pat = qr/F(o+)B(a+)r/;
my $str = "$pat";
$str =~ s/(?<!\\)(\(\?\w*)-([^i:]*)i([^i:]*):/$1i-$2$3:/g;
$pati = qr/$str/;
UPDATE: perl 5.14 quotes regexps in a different way, so my sample should probably look like
my $pat = qr/F(o+)B(a+)r/;
my $str = "$pat";
$str =~ s/(?<!\\)\(\?\^/(?^i/g;
$pati = qr/$str/;
But I don't have perl 5.14 at hand and can't test it.
UPD2: I also failed to check for escaped opening parenthesis.
I am matching a string of the form A<=>B!C<=>D!E<=>F... and want to do checks on the letters. Basically I want to tell if the letters are in the class according to a hash I have defined. I had the idea of doing the following regex and then looping through the matched strings:
$a =~ /(.)<=>(.)/g;
But I can't figure out to tell how many $1, $2 variables have matched. How do I know how many there are? Also, is there a better way to do this? I am using Perl 5.8.8.
You'll want the 'countof' operator to count the number of matches:
my $count = () = $string =~ /(.)<=>(.)/g;
Replacing the empty list with an array will retain the matches:
my #matches = $string =~ /(.)<=>(.)/g;
Which provides another way to get the $count:
my $count = #matches; # scalar #matches works too
Use a while loop
use warnings;
use strict;
my %letters = map { $_ => 1 } qw(A C F);
my $s = 'A<=>B!C<=>D!E<=>F';
while ($s =~ /(.)<=>(.)/g) {
print "$1\n" if exists $letters{$1};
print "$2\n" if exists $letters{$2};
}
__END__
A
C
F
Create a variable and increment it each time you go through your loop?
I want to be able to do a regex match on a variable and assign the results to the variable itself. What is the best way to do it?
I want to essentially combine lines 2 and 3 in a single line of code:
$variable = "some string";
$variable =~ /(find something).*/;
$variable = $1;
Is there a shorter/simpler way to do this? Am I missing something?
my($variable) = "some string" =~ /(e\s*str)/;
This works because
If the /g option is not used, m// in list context returns a list consisting of the subexpressions matched by the parentheses in the pattern, i.e., ($1, $2, $3 …).
and because my($variable) = ... (note the parentheses around the scalar) supplies list context to the match.
If the pattern fails to match, $variable gets the undefined value.
Why do you want it to be shorter? Does is really matter?
$variable = $1 if $variable =~ /(find something).*/;
If you are worried about the variable name or doing this repeatedly, wrap the thing in a subroutine and forget about it:
some_sub( $variable, qr/pattern/ );
sub some_sub { $_[0] = $1 if eval { $_[0] =~ m/$_[1]/ }; $1 };
However you implement it, the point of the subroutine is to make it reuseable so you give a particular set of lines a short name that stands in their place.
Several other answers mention a destructive substitution:
( my $new = $variable ) =~ s/pattern/replacement/;
I tend to keep the original data around, and Perl v5.14 has an /r flag that leaves the original alone and returns a new string with the replacement (instead of the count of replacements):
my $match = $variable =~ s/pattern/replacement/r;
Well, you could say
my $variable;
($variable) = ($variable = "find something soon") =~ /(find something).*/;
or
(my $variable = "find something soon") =~ s/^.*?(find something).*/$1/;
You can do substitution as:
$a = 'stackoverflow';
$a =~ s/(\w+)overflow/$1/;
$a is now "stack"
From Perl Cookbook 2nd ed
6.1 Copying and Substituting Simultaneously
$dst = $src;
$dst =~ s/this/that/;
becomes
($dst = $src) =~ s/this/that/;
I just assumed everyone did it this way, amazed that no one gave this answer.
Almost ....
You can combine the match and retrieve the matched value with a substitution.
$variable =~ s/.*(find something).*/$1/;
AFAIK, You will always have to copy the value though, unless you do not care to clobber the original.
$variable2 = "stackoverflow";
(my $variable1) = ($variable2 =~ /stack(\w+)/);
$variable1 now equals "overflow".
I do this:
#!/usr/bin/perl
$target = "n: 123";
my ($target) = $target =~ /n:\s*(\d+)/g;
print $target; # the var $target now is "123"
Also, to amplify the accepted answer using the ternary operator to allow you to specify a default if there is no match:
my $match = $variable =~ /(*pattern*).*/ ? $1 : *defaultValue*;