Matching string in perl which is parsed with awk - regex

Question

I did next with awk:
awk '/secon/ {print $1}' vladtest.sh |cut -c2-6
How to match this string in perl loop on appropriate way:
my $var1= `awk '/secon/ {print $1}' vladtest.sh |cut -c2-6`
if ($var1 eq "secon"){
print "OK";
} else {
print "FALSE"
}

Answer 1

First of all, your problem is the fact that the result of `...` includes the newline. So $var1 is not "secon", but "secon\n". You could deal with that any number of ways - wrapping chomp(...) around the whole assignment to $var1, or changing the right side of the eq to include the "\n", or using a regex instead: if ($var1 =~ /^secon$/) works with or without the chomp, because $ matches at a string-final newline if present.
Second of all, you're using about two programs too many here:
my $var1= `awk '/secon/ {print $1}' vladtest.sh |cut -c2-6`
I mean,awk can do anything cut can do, and perl can do anything either can do. When I'm typing one-liners at the shell prompt, I'm a big fan of awk and cut, simply for their economy of expression; within their respective specialities, their command lines are generally shorter than the equivalent (even using perl -a etc). But since you're already in a Perl program, there's no reason to shell out to something else here.
Here's an example drop-in replacement for the above:
my $var1;
{
open(my $fh, '<', 'vladtest.sh');
while (<$fh>) {
$var1 .= substr(split(' ')[0],1,5)."\n" if /secon/;
}
}
But you don't need to go through all that if you just want to detect if there's a match.
{open(my $fh, '<', 'vladtest.sh');
print 'NOT ' unless grep { /^\s*\Ssecon/ } <$fh>;
say 'OK';} # or print "OK\n"; if you don't have 5.10 extensions enabled.

Answer 2

My awk's a bit rusty, but you're grabbing $1 then looking for characters 2 through 6 to match 'secon'. If any line matches this, then print "OK", otherwise "NOT OK". Here's a perl only solution:
use 5.10.0;
while (<>) {
if (/^\s*\Ssecon/) {
say 'OK';
exit 0;
}
}
say 'NOT OK';
exit 1;
The regexp is made up of:
^ match the start of each line, followed by
\s* zero or more whitespace characters, followed by
\S one non-whitespace character, followed by
secon the literal string that you're interested in
As soon as we have a line that matches, we can print 'OK' and exit. Otherwise we'll fall through the end of the loop and print 'NOT OK'.
The use 5.10.0; line is needed so you can use say().

Answer 3

There are better ways to implement this request, but follow your current idea , change the first line to resolve the issue.
my $var1= `awk '$1~/secon/ {print substr($1,2,5)}' vladtest.sh`