Linux Perl-一个打印其脚本名称的脚本？

Question

I am running a script named statusCron.pl .

I want when I run that script, that script can detect/print its name which is statusCron.pl . By that I can detect its process ID so that if it found itself already running on the background it will kill itself.

By this I can eliminate hardcoding and re-apply it to all of my scripts.

Additional : I dont want the directory to be included.

Answer 1

You can use File::Basename like this:

use File::Basename;
print basename(__FILE__), "\n";

Answer 2

You can get the name of running program in $0 variable and to avoid the directory name we use the module File::Basename to get only the basename.

    use File::Basename;
    my $name = basename($0);
    print $name;

man perlvar for other special variables

Answer 3

您可以通过编写以下命令来打印Perl脚本文件的名称和进程ID

printf("%s %d\n", $0 =~ m|([^\\/]+)$|, $$);

Answer 4

#!/usr/bin/perl
print $0;

$0 stores the name of the script