How to add a modifier to a quoted regular (qr) expression - regex

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.

Related

How to capture every match in a global regex substitution?

I realize it is possible to achieve this with a slight workaround, but I am hoping there is a simpler way (since I often make use of this type of expression).
Given the example string:
my $str = "An example: sentence!*"
A regex can be used to match each punctuation mark and capture them in an array.
Thereafter, I can simply repeat the regex and replace the matches as in the following code:
push (#matches, $1), while ($str =~ /([\*\!:;])/);
$str =~ s/([\*\!:;])//g;
Would it be possible to combine this into a single step in Perl where substitution occurs globally while also keeping tabs on the replaced matches?
You can embed code to run in your regular expression:
my #matches;
my $str = 'An example: sentence!*';
$str =~ s/([\*\!:;])(?{push #matches, $1})//g;
But with a match this simple, I'd just do the captures and substitution separately.
Yes, it's possible.
my #matches;
$str =~ s/[*!:;]/ push #matches, $&; "" /eg;
However, I'm not convinced that the above is faster or clearer than the following:
my #matches = $str =~ /[*!:;]/g;
$str =~ tr/*!:;//d;
Use:
my $str = "An example: sentence!*";
my #matches = $str =~ /([\*\!:;])/g;
say Dumper \#matches;
$str =~ tr/*!:;//d;
Output:
$VAR1 = [
':',
'!',
'*'
];
Is that what you're looking for ?
my ($str, #matches) = ("An example: sentence!*");
#first method :
($str =~ s/([\*\!:;])//g) && push(#matches, $1);
#second method :
push(#matches, $1) while ($str =~ s/([\*\!:;])//g);
Try:
my $str = "An example: sentence!*";
push(#mys, ($str=~m/([^\w\s])/g));
print join "\n", #mys;
Thanks.

Perl replace delimiters

I have CSV text like
1,2,3,{4,5,6,7,8},9,10,100
I want to replace the delimiter of fields between {}. The text should look like:
1,2,3,{4|5|6|7|8},9,10,100
I tried perl -0777 -pe 's/\{.*?,\}/|/g'
but nothing happens. What should I do instead?
This will do as you ask. It replaces all commas that are followed by a sequence of characters that are not braces { }, and then a closing brace
use strict;
use warnings;
use 5.010;
my $s = '1,2,3,{4,5,6,7,8},9,10,100';
$s =~ s/,(?=[^{}]*\})/|/g;
say $s;
output
1,2,3,{4|5|6|7|8},9,10,100
You can use the following regex with $1$2| replacement string:
(\{\s*|(?<!^)\G)(\d+),(?=[,0-9]*\})
Output:
1,2,3,{4|5|6|7|8},9,10,100
Sample code:
#!/usr/bin/perl
$txt = "1,2,3,{4,5,6,7,8},9,10,100";
$txt =~ s/(\{\s*|(?<!^)\G)(\d+),(?=[,0-9]*\})/$1$2|/g;
print $txt;
Here's a command line version for Perl 5.14 and greater.
perl -pe 's/([{][\d,]+[}])/$1 =~ s~,~|~gr/ge'
The /e means it's evaluating the replacement as a Perl expression and not the standard regex expression. That means that it is taking the value of the first capture ($1) and performing a substitution with return (/r) so as to avoid the error trying to modify the read-only value ($1).
You can try this:
$st = "1,2,3,{4,5,6,7,8},9,10,100";
if ( $st=~/\{(.*)\}/ ) {
$tr = $1;
$tr =~ s/,/|/g;
$st =~ s/\{*\}/{$tr}/;
print "$st \n"
}
Output:
1,2,3,{4,5,6,7,8{4|5|6|7|8},9,10,100

Use variable as RegEx pattern

I'd like to use a variable as a RegEx pattern for matching filenames:
my $file = "test~";
my $regex1 = '^.+\Q~\E$';
my $regex2 = '^.+\\Q~\\E$';
print int($file =~ m/$regex1/)."\n";
print int($file =~ m/$regex2/)."\n";
print int($file =~ m/^.+\Q~\E$/)."\n";
The result (or on ideone.com):
0
0
1
Can anyone explain to me how I can use a variable as a RegEx pattern?
As documentation says:
$re = qr/$pattern/;
$string =~ /foo${re}bar/; # can be interpolated in other patterns
$string =~ $re; # or used standalone
$string =~ /$re/; # or this way
So, use the qr quote-like operator.
You cannot use \Q in a single-quoted / non-interpolated string. It must be seen by the lexer.
Anyway, tilde isn’t a meta-character.
Add use regex "debug" and you will see what is actually happening.

Matching a regular expression multiple times with Perl

Noob question here. I have a very simple perl script and I want the regex to match multiple parts in the string
my $string = "ohai there. ohai";
my #results = $string =~ /(\w\w\w\w)/;
foreach my $x (#results){
print "$x\n";
}
This isn't working the way i want as it only returns ohai. I would like it to match and print out ohai ther ohai
How would i go about doing this.
Thanks
Would this do what you want?
my $string = "ohai there. ohai";
while ($string =~ m/(\w\w\w\w)/g) {
print "$1\n";
}
It returns
ohai
ther
ohai
From perlretut:
The modifier "//g" stands for global matching and allows the
matching operator to match within a
string as many times as possible.
Also, if you want to put the matches in an array instead you can do:
my $string = "ohai there. ohai";
my #matches = ($string =~ m/(\w\w\w\w)/g);
foreach my $x (#matches) {
print "$x\n";
}
Or you could do this
my $string = "ohai there. ohai";
my #matches = split(/\s/, $string);
foreach my $x (#matches) {
print "$x\n";
}
The split function in this case splits on spaces and prints
ohai
there.
ohai

Match regex and assign results in single line of code

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*;