In Perl regex, how can I break from /ge loop..?
Let's say the code is:
s/\G(foo)(bar)(;|$)/{ break if $3 ne ';'; print "$1\n"; '' }/ge;
...break here doesn't work, but it should illustrate what I mean.
Generally, I would write this as a while statement:
while( s/(foo)(bar)/$1/ ) {
# my code to determine if I should stop
if(something) {
last;
}
}
The caveat with this method is that your search/replace will start at the beginning each time, which may matter depending on your regex.
If you really wanted to do it in the regex, you could write a function that returns an unmodified string if you reached your end point, such as a count in this case:
my $count=0;
sub myfunc {
my ($string, $a, $b) = #_;
$count++;
if($count > 3) {
return $string;
}
return $a;
}
$mystring = "foobar foobar, foobar + foobar and foobar";
$mystring =~ s/((foo)(bar))/myfunc($1,$2,$3)/ge;
# result: $mystring => "foo foo, foo + foobar and foobar"
If I knew your specific case, I could probably provide a more helpful example.
You can use some experimental features to emulate a break statement, the Perl documentation for some of these features warn that they may change in future versions of Perl.
my $str = "abcdef";
my $stop = 0;
$str =~ s/(?(?{ $stop })(?!))(.)/ $stop = 1 if $1 ge "c"; "X" /ge;
print "$str\n";
This will print XXXdef.
A piece wise explanation:
(?(condition)yes-pattern) if the pattern in in condition matches then match yes-pattern, otherwise don't match anything.
(?{ code }) execute code, inside a conditional if the code is true execute the yes-pattern
(?!) will always fail to match, it's meaning is something like "Don't match nothing" and since 'nothing' can be matched at any point in a string it will fail.
So when $stop is true the pattern can never match, and when $stop is false it matches.
Related
I want to match a variable character in a given string, but from the end.
Ideas on how to do this action?
for example:
sub removeCharFromEnd {
my $string = shift;
my $char = shift;
if($string =~ m/$char/){ // I want to match the char, searching from the end, $doesn't work
print "success";
}
}
Thank you for your assistance.
There is no regex modifier that would force Perl regex engine to parse the string from right to left. Thus, the most convenient way to achieve that is via a negative lookahead:
m/$char(?!.*$char)/
The (?!.*$char) negative lookahead will require the absence (=will fail the match if found) of a $char after any 0+ chars other than linebreak chars (use s modifier if you are running the regex against a multiline string input).
The regex engine works from left to right.
You can use the natural greediness of quantifiers to reach the end of the string and find the last char with the backtracking mechanism:
if($string =~ m/.*\K$char/s) { ...
\K marks the position of the match result beginning.
Other ways:
you can also reverse the string and use your previous pattern.
you can search all occurrences and take the last item in the list
I'm having trouble understanding what you want. Your subroutine is called removeCharFromEnd, so perhaps you want to remove $char from $string if it appears at the end of the string
You can do that like this
sub removeCharFromEnd {
my ( $string, $char ) = #_;
if ( $string =~ s/$char\z// ) {
print "success";
}
$string;
}
Or perhaps you want to remove the last occurrence of $char wherever it is. You can do that with
s/.*\K$char//
The subroutine I have written returns the modified string, so you would have to assign the result to a variable to save it. You can write
my $s = 'abc';
$s = removeCharFromEnd($s, 'c');
say $s;
output
ab
If you just want to modify the string in place then you should write
$ARGV[0] =~ s/$char\z//
using whichever substitution you choose. Then you can do this
my $s = 'abc';
removeCharFromEnd($s, 'c');
say $s;
This produces the same output
To get Perl to search from the end of a string, reverse the string.
sub removeCharFromEnd {
my $string = reverse shift #_;
my $char = quotemeta reverse shift #_;
$string =~ s/$char//;
$string = reverse $string;
return $string;
}
print removeCharFromEnd(qw( abcabc b )), "\n";
print removeCharFromEnd(qw( abcdefabcdef c )), "\n";
print removeCharFromEnd(qw( !"/$%?&*!"/$%?&* $ )), "\n";
I am using the following code to search for a substring and print it out with a few characters before and after it. Somehow Perl takes issue with me using $1 and complains about
Use of uninitialized value $1 in concatenation (.) or string.
I cannot figure out why...can you?
use List::Util qw[min max];
my $word = "test";
my $lines = "this is just a test to find something out";
my $context = 3;
while ($lines =~ m/\b$word\b/g ) { # as long as pattern is found...
print "$word\ ";
print "$1";
print substr ($lines, max(pos($lines)-length($1)-$context, 0), length($1)+$context); # check: am I possibly violating any boundaries here
}
You have to capture $word into regex group $1 by using parentheses,
while ($lines =~ m/\b($word)\b/g)
When you use $1, you are asking the code to use the first captured group from the regex and since your regex doesn't have any, well, that variable won't exist.
You can either refer to the whole match with $& or you add a capture group to your regex and keep using $1.
i.e. Either:
use List::Util qw[min max];
my $word = "test";
my $lines = "this is just a test to find something out";
my $context = 3;
while ($lines =~ m/\b$word\b/g ) { # as long as pattern is found...
print "$word\ ";
print "$&";
print substr ($lines, max(pos($lines)-length($&)-$context, 0), length($&)+$context); # check: am I possibly violating any boundaries here
}
Or
use List::Util qw[min max];
my $word = "test";
my $lines = "this is just a test to find something out";
my $context = 3;
while ($lines =~ m/(\b$word\b)/g ) { # as long as pattern is found...
print "$word\ ";
print "$1";
print substr ($lines, max(pos($lines)-length($1)-$context, 0), length($1)+$context); # check: am I possibly violating any boundaries here
}
Note: It doesn't matter whether you use (\b$word\b) or (\b$word)\b or \b($word\b) or \b($word)\b here because \b is a 'string' of 0 length.
When you want to address a matched part in regex, put it in parenthes. Than you'll be able to address this mathced part via $1 variable (for first pair of parenthes), $2 (for the second pair) and so on.
The values $1, $2 and so on hold the strings found by capture groups. When a match is performed all of these variables are set to undef. The code in the question does not have any capture groups and hence $1 is never given a value, it is undefined.
Running the code below shows the effect. Initially $1, $2 and $3 are not defined. The first match sets $1 and $2 but not $3. The second match sets only $1 but not that $2 is cleared to be undefined. The third match has no capture groups and all three are undefined.
use strict;
use warnings;
sub show
{
printf "\$1: %s\n", (defined $1 ? $1 : "-undef-");
printf "\$2: %s\n", (defined $2 ? $2 : "-undef-");
printf "\$3: %s\n", (defined $3 ? $3 : "-undef-");
print "\n";
}
my $text = "abcdefghij";
show();
$text =~ m/ab(cd)ef(gh)ij/; # First match
show();
$text =~ m/ab(cd)efghij/; # Second match
show();
$text =~ m/abcdefghij/; # Third match
show();
$1 will have no value unless you are actually capturing something.
You can adjust your boundary collection method to using lookahead and lookbehinds.
use strict;
use warnings;
my $lines = "this is just a test to find something out";
my $word = "test";
my $extra = 10;
while ($lines =~ m/(?:(?<=(.{$extra}))|(.{0,$extra}))\b(\Q$word\E)\b(?=(.{0,$extra}))/gs ) {
my $pre = $1 // $2;
my $word = $3;
my $post = $4;
print "'...$pre<$word>$post...'\n";
}
Outputs:
'...is just a <test> to find s...'
What I mean is:
For example, a{3,} will match 'a' at least three times greedly. It may find five times, 10 times, etc. I need this number. I need this number for the rest of the code.
I can do the rest less efficiently without knowing it, but I thought maybe Perl has some built-in variable to give this number or is there some trick to get it?
Just capture it and use length.
if (/(a{3,})/) {
print length($1), "\n";
}
Use #LAST_MATCH_END and #LAST_MATCH_START
my $str = 'jlkjmkaaaaaamlmk';
$str =~ /a{3,}/;
say $+[0]-$-[0];
Output:
6
NB: This will work only with a one-character pattern.
Here's an idea (maybe this is what you already had?) assuming the pattern you're interested in counting has multiple characters and variable length:
capture the substring which matches the pattern{3,} subpattern
then match the captured substring globally against pattern (note the absence of the quantifier), and force a list context on =~ to get the number of matches.
Here's a sample code to illustrate this (where $patt is the subpattern you're interested in counting)
my $str = "some catbratmatrattatblat thing";
my $patt = qr/b?.at/;
if ($str =~ /some ((?:$patt){3,}) thing/) {
my $count = () = $1 =~ /$patt/g;
print $count;
...
}
Another (admittedly somewhat trivial) example with 2 subpatterns
my $str = "some catbratmatrattatblat thing 11,33,446,70900,";
my $patt1 = qr/b?.at/;
my $patt2 = qr/\d+,/;
if ($str =~ /some ((?:$patt1){3,}) thing ((?:$patt2){2,})/) {
my ($substr1, $substr2) = ($1, $2);
my $count1 = () = $substr1 =~ /$patt1/g;
my $count2 = () = $substr2 =~ /$patt2/g;
say "count1: " . $count1;
say "count2: " . $count2;
}
Limitation(s) of this approach:
Fails miserably with lookarounds. See amon's example.
If you have a pattern of type /AB{n,}/ where A and B are complex patterns, we can split the regex into multiple pieces:
my $string = "ABABBBB";
my $n = 3;
my $count = 0;
TRY:
while ($string =~ /A/gc) {
my $pos = pos $string; # remember position for manual backtracking
$count++ while $string =~ /\GB/g;
if ($count < $n) {
$count = 0;
pos($string) = $pos; # restore previous position
} else {
last TRY;
}
}
say $count;
Output: 4
However, embedding code into the regex to do the counting may be more desirable, as it is more general:
my $string = "ABABBBB";
my $count;
$string =~ /A(?{ $count = 0 })(?:B(?{ $count++ })){3,}/ and say $count;
Output: 4.
The downside is that this code won't run on older perls. (Code was tested on v14 & v16).
Edit: The first solution will fail if the B pattern backtracks, e.g. $B = qr/BB?/. That pattern should match the ABABBBB string three times, but the strategy will only let it match two times. The solution using embedded code allows proper backtracking.
I want to understand how can I do arithmetic on matched sub-patterns in perl regex.
This is just a sample code and I want to understand how can I use \1 (already matched sub-pattern. in this case - 7) to match pattern+1 (8)
my $y = 77668;
if($y =~ /(\d)\1(\d)\2\1+1/) #How to increment a previously
#matched sub-pattern and form a pattern?
{
print $y;
}
EDIT
From the answers, I see that pattern arithmetic is not possible.
This is what I want to achieve.
I want to form a regex which will match this pattern:
N-3N-2N-1NNN+1N+2N+3 (N = 3,4,5,6
Its possible via regex code blocks:
my $y = 77668;
if($y =~ /(\d)\1(\d)\2(??{$1+1})/ ) {
print $y;
}
In this snippet (??{ CODE }) returns another regex that must match, so this regex looks like "8" ($1+1). As a result, whole regex will match only if 5th digit is greather and 1st by 1. But drawback with 1st digit is 9, this code block will return "10", so possible its wrong behavior, but you said nothing about what must be done in this case.
Now about N-3N-2N-1NNN+1N+2N+3 question, you can match it with this regex:
my $n = 5;
if( $y =~ /(??{ ($n-3).($n-2).($n-1).$n.($n+1).($n+2).($n+3) })/ ){
Or more "scalable" way:
my $n = 5;
if( $y =~ /(??{ $s=''; $s .= $n+$_ foreach(-3..3); $s; })/ ){
Again, what we must do if $n == 2 ?? $n-3 will be -1. Its not a simply digit cus it have sign, so you should think about this cases.
One another way. Match what we have and then check it.
if( $y =~ /(\d)(\d)(\d)(\d)(\d)(\d)(\d)/ ) {
if( $1 == ($4-3) && $2 == ($4-2) && $3 == ($4-1) && $6 == ($4+1) && $7 == ($4+2) && $7 == ($4+3) ) {
#...
Seems this method litle bit clumsy, but its obivious to everyone (i hope).
Also, you can optimize your regex since 7 ascending digits streak is not so frequent combination, plus get some lulz from co-workers xD:
sub check_number {
my $i;
for($i=1; $i<length($^N); $i++) {
last if substr($^N, $i, 1)<=substr($^N, $i-1, 1);
}
return $i<length($^N) ? "(*FAIL)" : "(*ACCEPT)";
}
if( $y =~ /[0123][1234][2345][3456][4567][5678][6789](??{ check_number() })/ ) {
Or... Maybe most human-friendly method:
if( $y =~ /0123456|1234567|2345678|3456789/ ) {
Seems last variant is bingo xD Its good example about not searching regex when things are so simple)
Of course this is possible. We are talking about Perl regexes after all. But it will be rather ugly:
say "55336"=~m{(\d)\1(\d)\2(\d)(?(?{$1+1==$3})|(*F))}?"match":"fail";
or pretty-printed:
say "55336" =~ m{ (\d)\1 (\d)\2 (\d)
(? (?{$1+1==$3}) # true-branch: nothing
|(*FAIL)
)
}x
? "match" : "fail";
What does this do? We collect the digits in ordinary captures. At the end, we use an if-else pattern:
(? (CONDITION) TRUE | FALSE )
We can embed code into a regex with (?{ code }). The return value of this code can be used as a condition. The (*FAIL) (short: (*F)) verb causes the match to fail. Use (*PRUNE) if you only want a branch, not the whole pattern to fail.
Embedded code is also great for debugging. However, older perls cannot use regexes inside this regex code :-(
So we can match lots of stuff and test it for validity inside the pattern itself. However, it might be a better idea to do that outside of the pattern like:
"string" =~ /regex/ and (conditions)
Now to your main pattern N-3N-2N-1NNN+1N+2N+3 (I hope I parsed it correctly):
my $super_regex = qr{
# N -3 N-2 N-1 N N N+1 N+2 N+3
(\d)-3\1-2\1-1\1\1(\d)(\d)(\d)
(?(?{$1==$2-1 and $1==$3-2 and $1==$4-3})|(*F))
}x;
say "4-34-24-144567" =~ $super_regex ? "match" : "fail";
Or did you mean
my $super_regex = qr{
#N-3 N-2 N-1 N N N+1 N+2 N+3
(\d)(\d)(\d) (\d)\4 (\d)(\d)(\d)
(? (?{$1==$4-3 and $2==$4-2 and $3==$4-1 and
$5==$4+1 and $6==$4+2 and $7==$4+3})|(*F))
}x;
say "123445678" =~ $super_regex ? "match" : "fail";
The scary thing is that these even works (with perl 5.12).
We could also generate parts of the pattern at match-time with the (??{ code }) construct — the return value of this code is used as a pattern:
my $super_regex = qr{(\d)(??{$1+1})(??{$1+2})}x;
say "234"=~$super_regex ? "match":"fail"
et cetera. However, I think readability suffers more this way.
If you need more than nine captures, you can use named captures with the
(?<named>pattern) ... \k<named>
constructs. The contents are also available in the %+ hash, see perlvar for that.
To dive further into the secrets of Perl regexes, I recommend reading perlre a few times.
here is my input string:
bar somthing foo bar somthing foo
I would like to count number of a character (ex: 't') between bar & foo
bar somthing foo -> 1
bar somthing foo -> 1
I know we can use /bar(.*?)foo/ and then count number of character in matches[1] with a String function
Is there way to do this w/o string function to count?
A Perl solution:
$_ = 'bar test this thing foo';
my $count = /bar(.*?)foo/ && $1 =~ tr/t//;
print $count;
Output:
4
Just for fun, using a single expression with (?{ code }):
$_ = 'bar test this thing foo';
my $count = 0;
/bar ( (?:(?!bar)[^t])*+ ) (?:t (?{ ++$count; }) (?-1) )*+ foo/x or $count = 0;
print $count;
#Qtax: the subject says PCRE... so it's not exactly perl. Hence (?{ code }) would most probably be unsupported (let alone the full perl code).
Though both solutions are cool ;)
#tqwer: you can get the match and then replace [^t] with "" and check length..
though i am not sure what the logic behind counting with regex would be ;)