假设您已经安装了:bash、pwd、dirname、echo;relpath是

#!/bin/bash
s=$(cd ${1%%/};pwd); d=$(cd $2;pwd); b=; while [ "${d#$s/}" == "${d}" ]
do s=$(dirname $s);b="../${b}"; done; echo ${b}${d#$s/}

我从pini和其他一些想法中得到了答案

注意:这要求两个路径都是现有文件夹。文件将无法工作。

这是我的版本。这是基于@Offirmo的回答。我使它与dash兼容，并修复了以下测试用例失败:

sh - compute-relative。“a / b / c - de - f / g " / a / b / c / def / g /" --> "../.. f - g - "

Now:

CT_FindRelativePath”a / b / c - de - f / g " / a / b / c / def / g /" --> "../../../ def - g”

查看代码:

# both $1 and $2 are absolute paths beginning with /
# returns relative path to $2/$target from $1/$source
CT_FindRelativePath()
{
    local insource=$1
    local intarget=$2

    # Ensure both source and target end with /
    # This simplifies the inner loop.
    #echo "insource : \"$insource\""
    #echo "intarget : \"$intarget\""
    case "$insource" in
        */) ;;
        *) source="$insource"/ ;;
    esac

    case "$intarget" in
        */) ;;
        *) target="$intarget"/ ;;
    esac

    #echo "source : \"$source\""
    #echo "target : \"$target\""

    local common_part=$source # for now

    local result=""

    #echo "common_part is now : \"$common_part\""
    #echo "result is now      : \"$result\""
    #echo "target#common_part : \"${target#$common_part}\""
    while [ "${target#$common_part}" = "${target}" -a "${common_part}" != "//" ]; do
        # no match, means that candidate common part is not correct
        # go up one level (reduce common part)
        common_part=$(dirname "$common_part")/
        # and record that we went back
        if [ -z "${result}" ]; then
            result="../"
        else
            result="../$result"
        fi
        #echo "(w) common_part is now : \"$common_part\""
        #echo "(w) result is now      : \"$result\""
        #echo "(w) target#common_part : \"${target#$common_part}\""
    done

    #echo "(f) common_part is     : \"$common_part\""

    if [ "${common_part}" = "//" ]; then
        # special case for root (no common path)
        common_part="/"
    fi

    # since we now have identified the common part,
    # compute the non-common part
    forward_part="${target#$common_part}"
    #echo "forward_part = \"$forward_part\""

    if [ -n "${result}" -a -n "${forward_part}" ]; then
        #echo "(simple concat)"
        result="$result$forward_part"
    elif [ -n "${forward_part}" ]; then
        result="$forward_part"
    fi
    #echo "result = \"$result\""

    # if a / was added to target and result ends in / then remove it now.
    if [ "$intarget" != "$target" ]; then
        case "$result" in
            */) result=$(echo "$result" | awk '{ string=substr($0, 1, length($0)-1); print string; }' ) ;;
        esac
    fi

    echo $result

    return 0
}

自2001年以来，它被内置到Perl中，因此它几乎可以在您能想象到的所有系统上工作，甚至VMS。

perl -le 'use File::Spec; print File::Spec->abs2rel(@ARGV)' FILE BASE

而且，解决方案很容易理解。

举个例子:

perl -le 'use File::Spec; print File::Spec->abs2rel(@ARGV)' $absolute $current

.．.会很好。

可悲的是，Mark Rushakoff的答案(现在删除了-它引用了这里的代码)在适应时似乎不能正确工作:

source=/home/part2/part3/part4
target=/work/proj1/proj2

注释中概述的思想可以经过改进，使其在大多数情况下正确工作。我假设脚本有一个源参数(您所在的位置)和一个目标参数(您想要到达的位置)，这两个参数要么是绝对路径名，要么是相对路径名。如果一个是绝对的，另一个是相对的，最简单的方法是用当前工作目录作为相对名称的前缀——但是下面的代码没有这样做。

当心

下面的代码接近正确工作，但不是很正确。

丹尼斯·威廉姆森(Dennis Williamson)的评论中提到了一个问题。 还有一个问题，这种纯文本的路径名处理，你可能会被奇怪的符号链接搞得一团糟。 该代码不处理'xyz/./pqr'等路径中的零散'点'。 该代码不处理'xyz/../pqr'等路径中的流浪'双点'。 简单地说:代码没有删除前导'。/'从路径。

Dennis的代码更好，因为它修复了1和5，但也有相同的问题2,3,4。 因此，请使用Dennis的代码(并在此之前对其进行投票)。

注意:POSIX提供了一个系统调用realpath()来解析路径名，这样路径名中就没有符号链接了。将其应用于输入名称，然后使用Dennis的代码每次都会给出正确的答案。编写包装realpath()的C代码很简单——我已经做过了——但我不知道有什么标准实用程序可以这样做。)

为此，我发现Perl比shell更容易使用，尽管bash对数组有很好的支持，并且可能也可以做到这一点——这是读者的练习。因此，给定两个兼容的名称，将它们分别分解为组件:

Set the relative path to empty. While the components are the same, skip to the next. When corresponding components are different or there are no more components for one path: If there are no remaining source components and the relative path is empty, add "." to the start. For each remaining source component, prefix the relative path with "../". If there are no remaining target components and the relative path is empty, add "." to the start. For each remaining target component, add the component to the end of the path after a slash.

因此:

#!/bin/perl -w

use strict;

# Should fettle the arguments if one is absolute and one relative:
# Oops - missing functionality!

# Split!
my(@source) = split '/', $ARGV[0];
my(@target) = split '/', $ARGV[1];

my $count = scalar(@source);
   $count = scalar(@target) if (scalar(@target) < $count);
my $relpath = "";

my $i;
for ($i = 0; $i < $count; $i++)
{
    last if $source[$i] ne $target[$i];
}

$relpath = "." if ($i >= scalar(@source) && $relpath eq "");
for (my $s = $i; $s < scalar(@source); $s++)
{
    $relpath = "../$relpath";
}
$relpath = "." if ($i >= scalar(@target) && $relpath eq "");
for (my $t = $i; $t < scalar(@target); $t++)
{
    $relpath .= "/$target[$t]";
}

# Clean up result (remove double slash, trailing slash, trailing slash-dot).
$relpath =~ s%//%/%;
$relpath =~ s%/$%%;
$relpath =~ s%/\.$%%;

print "source  = $ARGV[0]\n";
print "target  = $ARGV[1]\n";
print "relpath = $relpath\n";

测试脚本(方括号包含一个空格和一个制表符):

sed 's/#.*//;/^[    ]*$/d' <<! |

/home/part1/part2 /home/part1/part3
/home/part1/part2 /home/part4/part5
/home/part1/part2 /work/part6/part7
/home/part1       /work/part1/part2/part3/part4
/home             /work/part2/part3
/                 /work/part2/part3/part4

/home/part1/part2 /home/part1/part2/part3/part4
/home/part1/part2 /home/part1/part2/part3
/home/part1/part2 /home/part1/part2
/home/part1/part2 /home/part1
/home/part1/part2 /home
/home/part1/part2 /

/home/part1/part2 /work
/home/part1/part2 /work/part1
/home/part1/part2 /work/part1/part2
/home/part1/part2 /work/part1/part2/part3
/home/part1/part2 /work/part1/part2/part3/part4

home/part1/part2 home/part1/part3
home/part1/part2 home/part4/part5
home/part1/part2 work/part6/part7
home/part1       work/part1/part2/part3/part4
home             work/part2/part3
.                work/part2/part3

home/part1/part2 home/part1/part2/part3/part4
home/part1/part2 home/part1/part2/part3
home/part1/part2 home/part1/part2
home/part1/part2 home/part1
home/part1/part2 home
home/part1/part2 .

home/part1/part2 work
home/part1/part2 work/part1
home/part1/part2 work/part1/part2
home/part1/part2 work/part1/part2/part3
home/part1/part2 work/part1/part2/part3/part4

!

while read source target
do
    perl relpath.pl $source $target
    echo
done

测试脚本的输出:

source  = /home/part1/part2
target  = /home/part1/part3
relpath = ../part3

source  = /home/part1/part2
target  = /home/part4/part5
relpath = ../../part4/part5

source  = /home/part1/part2
target  = /work/part6/part7
relpath = ../../../work/part6/part7

source  = /home/part1
target  = /work/part1/part2/part3/part4
relpath = ../../work/part1/part2/part3/part4

source  = /home
target  = /work/part2/part3
relpath = ../work/part2/part3

source  = /
target  = /work/part2/part3/part4
relpath = ./work/part2/part3/part4

source  = /home/part1/part2
target  = /home/part1/part2/part3/part4
relpath = ./part3/part4

source  = /home/part1/part2
target  = /home/part1/part2/part3
relpath = ./part3

source  = /home/part1/part2
target  = /home/part1/part2
relpath = .

source  = /home/part1/part2
target  = /home/part1
relpath = ..

source  = /home/part1/part2
target  = /home
relpath = ../..

source  = /home/part1/part2
target  = /
relpath = ../../../..

source  = /home/part1/part2
target  = /work
relpath = ../../../work

source  = /home/part1/part2
target  = /work/part1
relpath = ../../../work/part1

source  = /home/part1/part2
target  = /work/part1/part2
relpath = ../../../work/part1/part2

source  = /home/part1/part2
target  = /work/part1/part2/part3
relpath = ../../../work/part1/part2/part3

source  = /home/part1/part2
target  = /work/part1/part2/part3/part4
relpath = ../../../work/part1/part2/part3/part4

source  = home/part1/part2
target  = home/part1/part3
relpath = ../part3

source  = home/part1/part2
target  = home/part4/part5
relpath = ../../part4/part5

source  = home/part1/part2
target  = work/part6/part7
relpath = ../../../work/part6/part7

source  = home/part1
target  = work/part1/part2/part3/part4
relpath = ../../work/part1/part2/part3/part4

source  = home
target  = work/part2/part3
relpath = ../work/part2/part3

source  = .
target  = work/part2/part3
relpath = ../work/part2/part3

source  = home/part1/part2
target  = home/part1/part2/part3/part4
relpath = ./part3/part4

source  = home/part1/part2
target  = home/part1/part2/part3
relpath = ./part3

source  = home/part1/part2
target  = home/part1/part2
relpath = .

source  = home/part1/part2
target  = home/part1
relpath = ..

source  = home/part1/part2
target  = home
relpath = ../..

source  = home/part1/part2
target  = .
relpath = ../../..

source  = home/part1/part2
target  = work
relpath = ../../../work

source  = home/part1/part2
target  = work/part1
relpath = ../../../work/part1

source  = home/part1/part2
target  = work/part1/part2
relpath = ../../../work/part1/part2

source  = home/part1/part2
target  = work/part1/part2/part3
relpath = ../../../work/part1/part2/part3

source  = home/part1/part2
target  = work/part1/part2/part3/part4
relpath = ../../../work/part1/part2/part3/part4

面对奇怪的输入，这个Perl脚本在Unix上运行得相当彻底(它没有考虑Windows路径名的所有复杂性)。它使用模块Cwd及其函数realpath解析存在的名称的真实路径，并对不存在的路径进行文本分析。在所有情况下，除了一种情况，它产生的输出都与Dennis的脚本相同。越轨的情况是:

source   = home/part1/part2
target   = .
relpath1 = ../../..
relpath2 = ../../../.

这两个结果是等价的，只是不完全相同。(输出来自测试脚本的一个轻微修改版本——下面的Perl脚本只是输出答案，而不是像上面的脚本那样输出输入和答案。)现在，我应该排除无效的答案吗?也许……

#!/bin/perl -w
# Based loosely on code from: http://unix.derkeiler.com/Newsgroups/comp.unix.shell/2005-10/1256.html
# Via: http://stackoverflow.com/questions/2564634

use strict;

die "Usage: $0 from to\n" if scalar @ARGV != 2;

use Cwd qw(realpath getcwd);

my $pwd;
my $verbose = 0;

# Fettle filename so it is absolute.
# Deals with '//', '/./' and '/../' notations, plus symlinks.
# The realpath() function does the hard work if the path exists.
# For non-existent paths, the code does a purely textual hack.
sub resolve
{
    my($name) = @_;
    my($path) = realpath($name);
    if (!defined $path)
    {
        # Path does not exist - do the best we can with lexical analysis
        # Assume Unix - not dealing with Windows.
        $path = $name;
        if ($name !~ m%^/%)
        {
            $pwd = getcwd if !defined $pwd;
            $path = "$pwd/$path";
        }
        $path =~ s%//+%/%g;     # Not UNC paths.
        $path =~ s%/$%%;        # No trailing /
        $path =~ s%/\./%/%g;    # No embedded /./
        # Try to eliminate /../abc/
        $path =~ s%/\.\./(?:[^/]+)(/|$)%$1%g;
        $path =~ s%/\.$%%;      # No trailing /.
        $path =~ s%^\./%%;      # No leading ./
        # What happens with . and / as inputs?
    }
    return($path);
}

sub print_result
{
    my($source, $target, $relpath) = @_;
    if ($verbose)
    {
        print "source  = $ARGV[0]\n";
        print "target  = $ARGV[1]\n";
        print "relpath = $relpath\n";
    }
    else
    {
        print "$relpath\n";
    }
    exit 0;
}

my($source) = resolve($ARGV[0]);
my($target) = resolve($ARGV[1]);
print_result($source, $target, ".") if ($source eq $target);

# Split!
my(@source) = split '/', $source;
my(@target) = split '/', $target;

my $count = scalar(@source);
   $count = scalar(@target) if (scalar(@target) < $count);
my $relpath = "";
my $i;

# Both paths are absolute; Perl splits an empty field 0.
for ($i = 1; $i < $count; $i++)
{
    last if $source[$i] ne $target[$i];
}

for (my $s = $i; $s < scalar(@source); $s++)
{
    $relpath = "$relpath/" if ($s > $i);
    $relpath = "$relpath..";
}
for (my $t = $i; $t < scalar(@target); $t++)
{
    $relpath = "$relpath/" if ($relpath ne "");
    $relpath = "$relpath$target[$t]";
}

print_result($source, $target, $relpath);

该脚本仅对绝对路径或没有绝对路径的相对路径的输入提供正确的结果。或者. .:

#!/bin/bash

# usage: relpath from to

if [[ "$1" == "$2" ]]
then
    echo "."
    exit
fi

IFS="/"

current=($1)
absolute=($2)

abssize=${#absolute[@]}
cursize=${#current[@]}

while [[ ${absolute[level]} == ${current[level]} ]]
do
    (( level++ ))
    if (( level > abssize || level > cursize ))
    then
        break
    fi
done

for ((i = level; i < cursize; i++))
do
    if ((i > level))
    then
        newpath=$newpath"/"
    fi
    newpath=$newpath".."
done

for ((i = level; i < abssize; i++))
do
    if [[ -n $newpath ]]
    then
        newpath=$newpath"/"
    fi
    newpath=$newpath${absolute[i]}
done

echo "$newpath"

假设您已经安装了:bash、pwd、dirname、echo;relpath是

#!/bin/bash
s=$(cd ${1%%/};pwd); d=$(cd $2;pwd); b=; while [ "${d#$s/}" == "${d}" ]
do s=$(dirname $s);b="../${b}"; done; echo ${b}${d#$s/}

我从pini和其他一些想法中得到了答案

注意:这要求两个路径都是现有文件夹。文件将无法工作。