This is my attempt to cut through extraneous issues raised "Why don’t my system calls work in the Perl program I wrap with pp?" I have created a simple Perl script on a linux system:
new-net:~/scripts # cat ls_test.pl
@ls_out = `ls -l`;
map { print "$_\n" } @ls_out;
$out = `sh out_test.sh`;
print "$out\n";
This script calls a simple shell file:
new-net:~/scripts # cat out_test.sh
echo "I'm here"
I used pp to pack the Perl script along with the shell script into ls_test:
new-net:~/test # unzip -l ls_test
Archive: ls_test
Length Date Time Name
-------- ---- ---- ----
0 07-13-09 16:41 script/
436 07-13-09 16:41 MANIFEST
214 07-13-09 16:41 META.yml
93 07-13-09 16:41 script/ls_test.pl
538 07-13-09 16:41 script/main.pl
16 07-13-09 16:20 out_test.sh
-------- -------
1297 6 files
If I run the packed file in an otherwise empty directory, the shell script is not found:
new-net:~/test # ./ls_test total 3391 -rwxr-xr-x 1 root root 3466177 Jul 13 16:41 ls_test sh: out_test.sh: No such file or directory
If I copy the shell script into the directory, the packed script runs as expected:
new-net:~/test # ./ls_test
total 3395 -rwxr-xr-x 1 root root 3466177 Jul 13 16:41 ls_test -rw-r--r-- 1 root root 16 Jul 13 16:20 out_test.sh I'm here
So, where does a pp packed script expect to find an included file? And how should a call to that included file be configured in the original Perl script?
The files in a packaged executable are extracted to a temporary directory (usually /tmp/par-USERNAME/cache-XXXXXXX). To access these files do something like the following:
#!/usr/bin/perl
# Reads a data file from the archive (added with -a)
print PAR::read_file("data");
# Will execute "script2" in the archive and exit. Will not return to this script.
require PAR;
PAR->import( { file => $0, run => 'script2' } );
You can also make symolic links to the executable that have the same name as the script you want to run, and run those.
Actually, rereading your question, simply accessing the PAR_TEMP environment variable is probably more useful:
#!/usr/bin/perl
use File::Slurp qw(slurp);
$data_dir = "$ENV{PAR_TEMP}/inc";
$script_dir = "$data_dir/script";
print slurp("$data_dir/datafile");
# file access permissions are not preserved as far as I can tell,
# so you'll have to invoke the interpreter explicitly.
system 'perl', "$script_dir/script2", @args;