Accessing shell variable in a Perl program
问题 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. 回答1: The variable $ddd isn't set *in the shell that you invoke from your Perl script. Ordinary shell variables are not