When I run this, it will say something "Can't exec ... No such file or directory...". That is because of the presence of "$1" in the bash script.

No. The first argument of (a multi-argument call to) system is the name of a command to execute. You're passing the entire script as the name of the command. You could run sh and pass the script to sh via the -c argument. 

#!/usr/bin/perl

use strict;
use warnings;

my $sh_script = do { local $/; <DATA> };

my @args = ( 'option1' );

system(sh => (-c => $sh_script, '--', @args) ) == 0
    or die "$?/$!\n";

__DATA__
#!/bin/sh

echo "hello"
echo $1
echo "hi"

[download]

Another option is to send the script to sh's stdin: 

...

open(my $sh_fh, '|-', sh => ('-s', '--', @args) )
    or die "$?/$!\n";
print $sh_fh $sh_script;
close($sh_fh);

...

[download]

The -- denotes the end of arguments destined for sh and the start arguments destined for the script.

Update: Added alternate method.

In reply to Re: Running Entire Bash Script Inside Perl by ikegami
in thread Running Entire Bash Script Inside Perl by bichonfrise74

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