I am completely new to perl and trying to design a lexer where I have come across:
my #token_def =
(
[Whitespace => qr{\s+}, 1],
[Comment => qr{#.*\n?$}m, 1],
);
and even after going through multiple sites I did not understand the meaning.
qr// is one of the quote-like operators that apply to pattern matching and related activities.
From perldoc:
This operator quotes (and possibly compiles) its STRING as a regular expression. STRING is interpolated the same way as PATTERN in m/PATTERN/. If ' is used as the delimiter, no interpolation is done.
From modern_perl:
The qr// operator creates first-class regexes. Interpolate them into the match operator to use them:
my $hat = qr/hat/;
say 'Found a hat!' if $name =~ /$hat/;
... or combine multiple regex objects into complex patterns:
my $hat = qr/hat/;
my $field = qr/field/;
say 'Found a hat in a field!'
if $name =~ /$hat$field/;
like( $name, qr/$hat$field/,
'Found a hat in a field!' );
qr// is documented in perlop in the "Regexp Quote-Like Operators" section.
Just like qq"..." aka "..." allows you to construct a string, qr/.../ allows you to construct a regular expression.
$s = "abc"; # Creates a string and assigns it to $s
$s = qq"abc"; # Same as above.
print("$s\n");
$re = qr/abc/; # Creates a compiled regex pattern and assigns it to $x
print "match\n" if $s =~ /$re/;
The quoting rules for qr/.../ are very similar to qq"..."'s. The only difference is that \ followed by a non-word character are passed through unchanged.
Related
Given :
my $str = "foo95285734776bar";
$str =~ s/([0-9]{2,4})/_????_/g;
What single regex where '????' is the length of $1 can produce output "foo_4__4__3_bar" ?
That is, where "9528" is replaced with "_4_", "5734" with "_4_", and the remaining "776" with "_3_".
You can use the /e modifier to add Perl code into the substitution part that is then evaled.
my $str = "foo95285734776bar";
$str =~ s/([0-9]{2,4})/'_' . length($1) . '_'/ge;
print $str;
Will output
foo_4__4__3_bar
Note that you now need a full Perl expression there. That's why you have to actually quote and concatenate the underscores.
From perlop:
A /e will cause the replacement portion to be treated as a full-fledged Perl expression and evaluated right then and there. It is, however, syntax checked at compile-time. A second e modifier will cause the replacement portion to be evaled before being run as a Perl expression.
I have a string of the following format:
word1.word2.word3
What are the ways to extract word2 from that string in perl?
I tried the following expression but it assigns 1 to sub:
#perleval $vars{sub} = $vars{string} =~ /.(.*)./; 0#
EDIT:
I have tried several suggestions, but still get the value of 1. I suspect that the entire expression above has a problem in addition to parsing. However, when I do simple assignment, I get the correct result:
#perleval $vars{sub} = $vars{string} ; 0#
assigns word1.word2.word3 to variable sub
. has a special meaning in regular expressions, so it needs to be escaped.
.* could match more than intended. [^.]* is safer.
The match operator (//) simply returns true/false in scalar context.
You can use any of the following:
$vars{sub} = $vars{string} =~ /\.([^.]*)\./ ? $1 : undef;
$vars{sub} = ( $vars{string} =~ /\.([^.]*)\./ )[0];
( $vars{sub} ) = $vars{string} =~ /\.([^.]*)\./;
The first one allows you to provide a default if there's no match.
Try:
/\.([^\.]+)\./
. has a special meaning and would need to be escaped. Then you would want to capture the values between the dots, so use a negative character class like ([^\.]+) meaning at least one non-dot. if you use (.*) you will get:
word1.stuff1.stuff2.stuff3.word2 to result in:
stuff1.stuff2.stuff3
But maybe you want that?
Here is my little example, I do find the perl one liners a little harder to read at times so I break it out:
use strict;
use warnings;
if ("stuff1.stuff2.stuff3" =~ m/\.([^.]+)\./) {
my $value = $1;
print $value;
}
else {
print "no match";
}
result
stuff2
. has a special meaning: any character (see the expression between your parentheses)
Therefore you have to escape it (\.) if you search a literal dot:
/\.(.*)\./
You've got to make sure you're asking for a list when you do the search.
my $x= $string =~ /look for (pattern)/ ;
sets $x to 1
my ($x)= $string =~ /look for (pattern)/ ;
sets $x to pattern.
How can I swap the letter o with the letter e and e with o?
I just tried this but I don't think this is a good way of doing this. Is there a better way?
my $str = 'Absolute force';
$str =~ s/e/___eee___/g;
$str =~ s/o/e/g;
$str =~ s/___eee___/o/g;
Output: Abseluto ferco
Use the transliteration operator:
$str =~ y/oe/eo/;
E.g.
$ echo "Absolute force" | perl -pe 'y/oe/eo/'
Abseluto ferco
As has already been said, the way to do this is the transliteration operator
tr/SEARCHLIST/REPLACEMENTLIST/cdsr
y/SEARCHLIST/REPLACEMENTLIST/cdsr
Transliterates all occurrences of the characters found in the search list with the corresponding character in the replacement list. It returns the number of characters replaced or deleted. If no string is specified via the =~ or !~ operator, the $_ string is transliterated.
However, I want to commend you on your creative use of regular expressions. Your solution works, although the placeholder string _ee_ would've been sufficient.
tr is only going to help you for character replacements though, so I'd like to quickly teach you how to utilize regular expressions for a more complicated mass replacement. Basically, you just use the /e tag to execute code in the RHS. The following will also do the replacement you were aiming for:
my $str = 'Absolute force';
$str =~ s/([eo])/$1 eq 'e' ? 'o' : 'e'/eg;
print $str;
Outputs:
Abseluto ferco
Note how the LHS (left hand side) matches both o and e, and them the RHS (right hand side) does a test to see which matched and returns the opposite for replacement.
Now, it's common to have a list of words that you want to replace, so it's convenient to just build a hash of your from/to values and then dynamically build the regular expression. The following does that:
my $str = 'Hello, foo. How about baz? Never forget bar.';
my %words = (
foo => 'bar',
bar => 'baz',
baz => 'foo',
);
my $wordlist_re = '(?:' . join('|', map quotemeta, keys %words) . ')';
$str =~ s/\b($wordlist_re)\b/$words{$1}/eg;
Outputs:
Hello, bar. How about foo? Never forget baz.
This above could've worked for your e and o case, as well, but would've been overkill. Note how I use quotemeta to escape the keys in case they contained a regular expression special character. I also intentionally used a non-capturing group around them in $wordlist_re so that variable could be dropped into any regex and behave as desired. I then put the capturing group inside the s/// because it's important to be able to see what's being captured in a regex without having to backtrack to the value of an interpolated variable.
The tr/// operator is best. However, if you wanted to use the s/// operator (to handle more than just single letter substitutions), you could write
$ echo 'Absolute force' | perl -pe 's/(e)|o/$1 ? "o" : "e"/eg'
Abseluto ferco
The capturing parentheses avoid the redundant $1 eq 'e' test in #Miller's answer.
from man sed:
y/source/dest/
Transliterate the characters in the pattern space which appear in source to the corresponding character in dest.
and tr command can do this too:
$ echo "Absolute force" | tr 'oe' 'eo'
Abseluto ferco
When I do this
#!/usr/bin/perl
use strict;
use warnings;
use Data::Dumper;
my $s = 'dfgdfg5 )';
my $a = '5 )';
my $b = '567';
$s =~ s/$a/$b/g;
print Dumper $s;
I get
Unmatched ) in regex; marked by <-- HERE in m/5 ) <-- HERE / at ./test.pl line 11.
The problem is that $a have a (.
How do I prevent the regex from failing?
Update
The string in $a do I get from a database query, so I can't change it. Or would it be possible to make an $a2 where "something" searches for ) and replaces them with \)?
You need to escape it. Either manually by adding backslash in front of it, or by using quotemeta or the \Q sequence inside the regex:
$a = quotemeta($a);
Or
$s =~ /\Q$a/$b/g;
ETA: This is a good option if you want to match literal strings from a database query.
You should also be aware that it is not a good idea to use $a and $b as variables, since they will mask the predefined variables that are used with sort. E.g. sort { $a <=> $b } #foo.
The simple answer is to backslash escape the paren. my $a = '5 \)'; In your case, as your post mentions, you aren't the one creating the strings, so literally escaping them isn't an option.
It may be simpler to just wrap the variable that's being interpolated by the regex inside of a \Q ... \E.
$s =~ s/\Q$a\E/$b/g;
The quotemeta() function may also be helpful to you, depending on how your code is factored. With that option you would pass $a through quotemeta before interpolating it in the regex. \Q...\E is probably easier in this situation, but if your code is simplified by using quotemeta instead, it's there for you.
Use \) instead of just ). ) is special because it's normally used for capturing patterns so you need to escape it first.
Escape the parentheses with a backslash:
my $a = '5 \)'oi;
Or use \Q inside the regexp:
$s =~ s/\Q$a/$b/g;
Also when storing regexps in a variable, you should look into the regexp quote operator: http://perldoc.perl.org/perlop.html#Regexp-Quote-Like-Operators
my $a = qr/5 \)/oi;
In Perl regular expression you need to mask special chars with a backslash \.
Try
my $a = '5 \)';
my $b = '567';
$s =~ s/$a/$b/g;
For details and a good start see perldoc perlretut
Update: I didn't know the RE came from a database. Well, the code above works nevertheless. The hint for the tutorial still applies.
I think you just need to escape the brackets, ie replace ) with \)
I'm using a Perl program to extract text from a file. I have an array of strings which I use as delimiters for the text, e.g:
$pat = $arr[1] . '(.*?)' . $arr[2];
if ( $src =~ /$pat/ ) {
print $1;
}
However, two of the strings in the array are $450 and (Buy now). The problem with these is that the symbols in the strings represent end-of-string and capture group in Perl regular expressions, so the text doesn't parse as I intend.
Is there a way around this?
Try Perl's quotemeta function. Alternatively, use \Q and \E in your regex to turn off interpolation of values in the regex. See perlretut for more on \Q and \E - they may not be what you're looking for.
quotemeta escapes meta-characters so they are interpreted as literals. As a shortcut, you can use \Q...\E in double-quotish context to surround stuff that should be quoted:
$pat = quotemeta($arr[1]).'(.*?)'.quotemeta($arr[2]);
if($src=~$pat) { print $1 }
or
$pat = "\Q$arr[1]\E(.*?)\Q$arr[2]"; # \E not necessary at the end
if($src=~$pat) { print $1 }
or just
if ( $src =~ /\Q$arr[1]\E(.*?)\Q$arr[2]/ ) { print $1 }
Note that this isn't limited to interpolated variables; literal characters are affected too:
perl -wle'print "\Q.+?"'
\.\+\?
though obviously it happens after variable interpolation, so "\Q$foo" doesn't become '\$foo'.
Use quotemeta:
$pat = quotemeta($arr[1]) . '(.*?)' . quotemeta($arr[2]);
if ($src =~ $pat)
print $1;