如何在Perl中获取目录（文件路径）分隔符？

Question

In case of Java, we can get the path separator using

System.getProperty("path.separator");

Is there a similar way in Perl? All I want to do is to find a dir, immediate sub directory. Say I am being given two arguments $a and $b ; I am splitting the first one based on the path separator and joining it again except the last fragment and comparing with the second argument.

The problem is my code has to be generic and for that I need to know whats the system dependent path separator is?

Answer 1

You should not form file paths by hand - instead use File::Spec module:

($volume, $directories,$file) = File::Spec->splitpath( $path );
@dirs = File::Spec->splitdir( $directories );
$path = File::Spec->catdir( @directories );
$path = File::Spec->catfile( @directories, $filename );

Answer 2

您可以在File :: Util中使用SL常量 。

Answer 3

The accepted answer solves your real problem, but if you really want to get the separator (using only perl core modules):

my $sep = File::Spec->catfile('', '');

This joins two empty file names with the current system's separator, leaving only the separator.