Perl search and replace with variable and capture group - regex

As the question says, I am trying to do a search replace using a variable and a capture group. That is, the replace string contains $1. I followed the answers here and here, but they did not working for me; $1 comes through in the replace. Can you help me spot my problem?
I am reading my regular expressions from a file like so:
while( my $line = <$file>) {
my #findRep = split(/:/, $line);
my $find = $findRep[0];
my $replace = '"$findRep[2]"'; # This contains the $1
$allTxt =~ s/$find/$replace/ee;
}
If I manually set my $replace = '"$1 stuff"' the replace works as expected. I have played around with every single/double quoting and /e combination I can think of.

You're using single quotes so $findRep[2] isn't interpolated. Try this instead:
my $replace = qq{"$findRep[2]"};

Why regex replacement when you already have your values in #findRep
while( my $line = <$file>) {
my #findRep = split(/:/, $line);
$findRep[0] = $findRep[2];
my $allTxt = join(":", #findRep);
}

Related

Applying Filters in Perl using Regex

I'm trying to extract text and numbers from a string using regex in perl. Here is my code:
$line = "finish=100\n";
($var) = $line =~ /[a-z]+/;
($val) = $line =~ /[0-9]+/;
My expected output is that $var = "finish" and $val = 100. However when I run the code $var = 1 and $val = 1.
Any help would be appreciated!!
Use capturing parentheses inside your regular expressions:
$line = "finish=100\n";
($var) = $line =~ /([a-z]+)/;
($val) = $line =~ /([0-9]+)/;
print "$var $val\n";
Refer to perlre
A regex match in list context (where the regex doesn't use the /g flag) returns
the empty list if it fails
a list of captured substrings ($1, $2, ...) if it succeeds and the pattern contains capturing groups
the list 1 if it succeeds and the pattern doesn't capture anything
Your regexes match, but they don't contain any capturing groups, so that's why you get 1 in $var and $val.
If you add capturing groups (/([a-z]+)/, /([0-9]+)/), you get the matched substrings instead.
Note that it might be easier to just do it all in one match:
my ($var, $val) = $line =~ /^([a-z]+)=([0-9]+)$/;
This way you also validate that the input string has the expected form and isn't just something like "Cat o' 9 tails", which (with your original regexes) would extract $var = "at" and $val = "9".
You can too get two values in one array, maybe with this:
$line = "finish=100\n";
#matches = $line =~ /(\w+)\W(\d+)/;
print "$matches[0], $matches[1]";

Extract string after a symbol in Perl

How can I extract string after a symbol in Perl?
I tried doing some searches but even the code I found didn't work.
I'm trying to extract the string after a colon. So I want to show everything after the colon.
Example:
string = day1: string over here
substring = string over here
So far I have tried:
$substring = $string=~ /(\:.*)\s*$/;
But it only outputs the number 1 over and over.
That's because pattern matches in a scalar context are boolean tests. If you want to capture bracket content (capture groups), you need a list context. It's ok if the list is only one element though:
try this:
my ( $substring ) = $string=~ /(\:.*)\s*$/;
Difference maybe a bit subtle, but basically - we are assigning 'all the hits' from the pattern match to a list... that comprises one element.
Note - that's so you can do:
my #matches = $string =~ m/(.)/g;
And get multiple 'hits' returned. If you do as above, you will only get the first match - which is irrelevant given your pattern, but you can do:
my ( $key, $value ) = $string =~ m/(\w+)=(\w+)/;
for example.
I usually use parentheses to extract a part from text and then refer to the result stored in $1 variable.
look at example:
my $text = "day1: string over here";
print $1 if ($text =~ /:\s*(.+)$/);
but similar result may be recieved with this code too:
my $text = "day1: string over here";
my ($a) = $text =~ /:\s*(.+)$/;
print $a;
You can achieve desire substring by using split function also:
#!/usr/bin/perl
use warnings;
use strict;
my $string = "day1: string over here";
my (undef, $substring) = split(':\s*', $string);
print $substring, "\n";
Output:
string over here
Or you can get this by using capturing group () in regex:
my $string = "day1: string over here";
$string =~ m/(.*)\:\s+(.*)$/;
my $substring = $2;
print $substring, "\n";

Perl how do you assign a varanble to a regex match result

How do you create a $scalar from the result of a regex match?
Is there any way that once the script has matched the regex that it can be assigned to a variable so it can be used later on, outside of the block.
IE. If $regex_result = blah blah then do something.
I understand that I should make the regex as non-greedy as possible.
#!/usr/bin/perl
use strict;
use warnings;
# use diagnostics;
use Win32::OLE;
use Win32::OLE::Const 'Microsoft Outlook';
my #Qmail;
my $regex = "^\\s\*owner \#";
my $sentence = $regex =~ "/^\\s\*owner \#/";
my $outlook = Win32::OLE->new('Outlook.Application')
or warn "Failed Opening Outlook.";
my $namespace = $outlook->GetNamespace("MAPI");
my $folder = $namespace->Folders("test")->Folders("Inbox");
my $items = $folder->Items;
foreach my $msg ( $items->in ) {
if ( $msg->{Subject} =~ m/^(.*test alert) / ) {
my $name = $1;
print " processing Email for $name \n";
push #Qmail, $msg->{Body};
}
}
for(#Qmail) {
next unless /$regex|^\s*description/i;
print; # prints what i want ie lines that start with owner and description
}
print $sentence; # prints ^\\s\*offense \ # not lines that start with owner.
One way is to verify a match occurred.
use strict;
use warnings;
my $str = "hello what world";
my $match = 'no match found';
my $what = 'no what found';
if ( $str =~ /hello (what) world/ )
{
$match = $&;
$what = $1;
}
print '$match = ', $match, "\n";
print '$what = ', $what, "\n";
Use Below Perl variables to meet your requirements -
$` = The string preceding whatever was matched by the last pattern match, not counting patterns matched in nested blocks that have been exited already.
$& = Contains the string matched by the last pattern match
$' = The string following whatever was matched by the last pattern match, not counting patterns matched in nested blockes that have been exited already. For example:
$_ = 'abcdefghi';
/def/;
print "$`:$&:$'\n"; # prints abc:def:ghi
The match of a regex is stored in special variables (as well as some more readable variables if you specify the regex to do so and use the /p flag).
For the whole last match you're looking at the $MATCH (or $& for short) variable. This is covered in the manual page perlvar.
So say you wanted to store your last for loop's matches in an array called #matches, you could write the loop (and for some reason I think you meant it to be a foreach loop) as:
my #matches = ();
foreach (#Qmail) {
next unless /$regex|^\s*description/i;
push #matches_in_qmail $MATCH
print;
}
I think you have a problem in your code. I'm not sure of the original intention but looking at these lines:
my $regex = "^\\s\*owner \#";
my $sentence = $regex =~ "/^\s*owner #/";
I'll step through that as:
Assign $regexto the string ^\s*owner #.
Assign $sentence to value of running a match within $regex with the regular expression /^s*owner $/ (which won't match, if it did $sentence will be 1 but since it didn't it's false).
I think. I'm actually not exactly certain what that line will do or was meant to do.
I'm not quite sure what part of the match you want: the captures, or something else. I've written Regexp::Result which you can use to grab all the captures etc. on a successful match, and Regexp::Flow to grab multiple results (including success statuses). If you just want numbered captures, you can also use Data::Munge
You can do the following:
my $str ="hello world";
my ($hello, $world) = $str =~ /(hello)|(what)/;
say "[$_]" for($hello,$world);
As you see $hello contains "hello".
If you have older perl on your system like me, perl 5.18 or earlier, and you use $ $& $' like codequestor's answer above, it will slow down your program.
Instead, you can use your regex pattern with the modifier /p, and then check these 3 variables: ${^PREMATCH}, ${^MATCH}, and ${^POSTMATCH} for your matching results.

perl regular expression finding pattern only in the front of the text

Suppose there is a text like this:
|-SAMPLE-D2
|---SAMPLE-D1
|---SAMPLE3
I want to count the number of "-" after |.
I tried to parse that by using the following regular expression in perl
$count=()= /-/g;
but this is problematic because the first two has "-" somewhere else in the text as well as in the front. How should I form my regex or use other function in perl to get the number of "-" right after "|"?
Regex to match the dashes after the starting |:
/^\|([\-]*)/
To count dashes that are not preceded by a letter, use a negative look-behind assertion.
$count = () = /(?<!\w)-/g
If the vertical line only ever comes at the start you can get the string of repeating minuses with:
my ($match) = $txt =~ /^\|(-*)/;
The brackets around $match cause the captured portion of the regex to be put into it
then get the number of minuses using
my $minus_count = length($match || '');
The
|| '')
bit
Initialises $match if the regex above found no matches at all, to stop length moaning about uninitialised variables (if you have warnings on)
Not sure if you can count in Regex directly but you can extract capture groups and do a simple arithmetic with their string lengths:
#!/usr/bin/perl
use warnings;
my $inFile = $ARGV[0];
open(FILEHANDLE, "<", $inFile) || die("Could not open file ".$inFile);
my #fileLines = <FILEHANDLE>;
my $lineNo = 0;
my $rslt;
foreach my $line(#fileLines) {
chomp($line);
$line =~ s/^\s+//;
$line =~ s/\s+$//;
$lineNo++;
print "\n".$lineNo." = <".$line.">";
if($line =~ m/^\|-+(.+)/) {
my $text = $1;
print "\n\ttext = <".$text.">";
my $minCnt = length($line) - length($text) - 1;
print "\n\tminus count = <".$minCnt.">";
}
}
close(FILEHANDLE);

Regex parsing string substitution question

I would like to know if there is an easy way for parsing a string like this
set PROMPT = Yes, Master?
What I would like to do, is parse one part of this string up to the equal sign and parse the second part after the equal sign into another string.
Something like...
$phrase = 'set PROMPT = Yes, Master?';
#parts = split /=/, $phrase;
or
($set, $value) = split /=/, $phrase, 2;
[updated] Changes per comments.
Try matching this regex /\s*set\s*(\w+)\s*=\s*(.*)\s*$/ and setting the parts with $1 and $2:
my $str = 'set PROMPT = Yes, Master?';
my ($k, $v) = ($1, $2) if $str =~ /\s*set\s*(\w+)\s*=\s*(.*)\s*$/;
print "OK: k=$k, v=$v\n"; OK: k=PROMPT, v=Yes, Master?
while ($subject =~ m/([^\s]+)\s*=\s*([^\$]+)/img) {
# $1 = $2
}