perl - 如何规范化 Perl 中的路径？ (不检查文件系统)

Question

我想要 Perl 相当于 Python 的 os.path.normpath() :

Normalize a pathname by collapsing redundant separators and up-level references so that A//B, A/B/, A/./B and A/foo/../B all become A/B. This string manipulation may change the meaning of a path that contains symbolic links. […]

例如，我想转换 '/a/../b/./c//d'进入 /b/c/d .

我正在操作的路径并不代表本地文件树中的真实目录。不涉及符号链接(symbolic link)。因此，简单的字符串操作可以正常工作。

我试过 Cwd::abs_path 和 File::Spec ，但他们不做我想做的事。

my $path = '/a/../b/./c//d';

File::Spec->canonpath($path);
File::Spec->rel2abs($path, '/');
# Both return '/a/../b/c/d'.
# They don't remove '..' because it might change
# the meaning of the path in case of symlinks.

Cwd::abs_path($path);
# Returns undef.
# This checks for the path in the filesystem, which I don't want.

Cwd::fast_abs_path($path);
# Gives an error: No such file or directory

可能相关链接:

Normalized directory paths - perlmonks : 人们讨论了几种方法。

Answer 1

鉴于 File::Spec 几乎是我所需要的，我最终编写了一个删除 ../ 的函数。来自 File::Spec->canonpath() . The full code including tests is available as a GitHub Gist .

use File::Spec;

sub path_normalize_by_string_manipulation {
    my $path = shift;

    # canonpath does string manipulation, but does not remove "..".
    my $ret = File::Spec->canonpath($path);

    # Let's remove ".." by using a regex.
    while ($ret =~ s{
        (^|/)              # Either the beginning of the string, or a slash, save as $1
        (                  # Followed by one of these:
            [^/]|          #  * Any one character (except slash, obviously)
            [^./][^/]|     #  * Two characters where
            [^/][^./]|     #    they are not ".."
            [^/][^/][^/]+  #  * Three or more characters
        )                  # Followed by:
        /\.\./             # "/", followed by "../"
        }{$1}x
    ) {
        # Repeat this substitution until not possible anymore.
    }

    # Re-adding the trailing slash, if needed.
    if ($path =~ m!/$! && $ret !~ m!/$!) {
        $ret .= '/';
    }

    return $ret;
}