I have this Perl script:
#!/usr/bin/perl
$var = `ls -l \$ddd` ;
print $var, "\n";
And ddd is a shell variable
$ echo "$ddd"
arraytest.pl
When I execute the Perl script I get a listing of all files in the directory instead of just one file, whose file name is contained in shell variable $ddd.
Whats happening here ? Note that I am escaping $ddd in backticks in the Perl script.
The variable $ddd isn't set *in the shell that you invoke from your Perl script.
Ordinary shell variables are not inherited by subprocesses. Environment variables are.
If you want this to work, you'll need to do one of the following in your shell before invoking your Perl script:
ddd=arraytest.pl ; export ddd # sh
export ddd=arraytest.pl # bash, ksh, zsh
setenv ddd arraytest.pl # csh, tcsh
This will make the environment variable $ddd visible from your Perl script. But then it probably makes more sense to refer to it as $ENV{ddd}, rather than passing the literal string '$ddd' to the shell and letting it expand it:
$var = `ls -l $ENV{ddd}`;
sh variant is of course portable to any Bourne-compatible shell (including ksh etc); the export variable=value syntax is ambiguous between shells so you should avoid it unless you are specifically targeting e.g. Bash. Details at unix.stackexchange.com/questions/193095/…You forgot to export ddd:
Mark each name to be passed to child processes in the environment.
So ddd is not automatically available to child processes.
%ENV because it is a shell variable, not an environment variable until you export it. Compare ddd='pancakes';perl -e 'say $ENV{ddd}, ddd='pancakes' perl -e 'say $ENV{ddd}', and export ddd='pancakes';perl -e 'say $ENV{ddd}'.The hash %ENV contains your current environment.
$var = `ls -l $ENV{ddd}`;
/edit - it works, checked, of course ddd need to be exported before running script
export ddd='arraytest.pl'
perl script.pl
'$ddd' to the shell, which would have expanded it.