$ python -c "import os.path; print os.path.relpath('/foo/bar', '/foo/baz/foo')"

给:

../../bar

这个脚本只对路径名有效。它不需要任何文件存在。如果传递的路径不是绝对的，那么行为就有点不寻常，但是如果两条路径都是相对的，那么应该能正常工作。

我只在OS X上测试过，所以可能不太便携。

#!/bin/bash
set -e
declare SCRIPT_NAME="$(basename $0)"
function usage {
    echo "Usage: $SCRIPT_NAME <base path> <target file>"
    echo "       Outputs <target file> relative to <base path>"
    exit 1
}

if [ $# -lt 2 ]; then usage; fi

declare base=$1
declare target=$2
declare -a base_part=()
declare -a target_part=()

#Split path elements & canonicalize
OFS="$IFS"; IFS='/'
bpl=0;
for bp in $base; do
    case "$bp" in
        ".");;
        "..") let "bpl=$bpl-1" ;;
        *) base_part[${bpl}]="$bp" ; let "bpl=$bpl+1";;
    esac
done
tpl=0;
for tp in $target; do
    case "$tp" in
        ".");;
        "..") let "tpl=$tpl-1" ;;
        *) target_part[${tpl}]="$tp" ; let "tpl=$tpl+1";;
    esac
done
IFS="$OFS"

#Count common prefix
common=0
for (( i=0 ; i<$bpl ; i++ )); do
    if [ "${base_part[$i]}" = "${target_part[$common]}" ] ; then
        let "common=$common+1"
    else
        break
    fi
done

#Compute number of directories up
let "updir=$bpl-$common" || updir=0 #if the expression is zero, 'let' fails

#trivial case (after canonical decomposition)
if [ $updir -eq 0 ]; then
    echo .
    exit
fi

#Print updirs
for (( i=0 ; i<$updir ; i++ )); do
    echo -n ../
done

#Print remaining path
for (( i=$common ; i<$tpl ; i++ )); do
    if [ $i -ne $common ]; then
        echo -n "/"
    fi
    if [ "" != "${target_part[$i]}" ] ; then
        echo -n "${target_part[$i]}"
    fi
done
#One last newline
echo

这里的答案并不是每天都能用的。由于在纯bash中很难正确地做到这一点，我建议以下可靠的解决方案(类似于注释中的一个建议):

function relpath() { 
  python -c "import os,sys;print(os.path.relpath(*(sys.argv[1:])))" "$@";
}

然后，你可以得到基于当前目录的相对路径:

echo $(relpath somepath)

或者你可以指定路径相对于给定的目录:

echo $(relpath somepath /etc)  # relative to /etc

一个缺点是这需要python，但是:

它在任何python >= 2.6中工作相同 它不要求文件或目录存在。 文件名可以包含更广泛的特殊字符。 例如，如果文件名包含 空格或其他特殊字符。 它是一个单行函数，不会使脚本混乱。

注意，包含basename或dirname的解决方案不一定更好，因为它们要求安装coreutils。如果有人有可靠而简单的纯bash解决方案(而不是令人费解的好奇心)，我会感到惊讶。

我猜这个也可以…(自带内置测试):)

好吧，预计会有一些开销，但我们在这里做的是伯恩壳!；）

#!/bin/sh

#
# Finding the relative path to a certain file ($2), given the absolute path ($1)
# (available here too http://pastebin.com/tWWqA8aB)
#
relpath () {
  local  FROM="$1"
  local    TO="`dirname  $2`"
  local  FILE="`basename $2`"
  local  DEBUG="$3"

  local FROMREL=""
  local FROMUP="$FROM"
  while [ "$FROMUP" != "/" ]; do
    local TOUP="$TO"
    local TOREL=""
    while [ "$TOUP" != "/" ]; do
      [ -z "$DEBUG" ] || echo 1>&2 "$DEBUG$FROMUP =?= $TOUP"
      if [ "$FROMUP" = "$TOUP" ]; then
        echo "${FROMREL:-.}/$TOREL${TOREL:+/}$FILE"
        return 0
      fi
      TOREL="`basename $TOUP`${TOREL:+/}$TOREL"
      TOUP="`dirname $TOUP`"
    done
    FROMREL="..${FROMREL:+/}$FROMREL"
    FROMUP="`dirname $FROMUP`"
  done
  echo "${FROMREL:-.}${TOREL:+/}$TOREL/$FILE"
  return 0
}

relpathshow () {
  echo " - target $2"
  echo "   from   $1"
  echo "   ------"
  echo "   => `relpath $1 $2 '      '`"
  echo ""
}

# If given 2 arguments, do as said...
if [ -n "$2" ]; then
  relpath $1 $2

# If only one given, then assume current directory
elif [ -n "$1" ]; then
  relpath `pwd` $1

# Otherwise perform a set of built-in tests to confirm the validity of the method! ;)
else

  relpathshow /usr/share/emacs22/site-lisp/emacs-goodies-el \
              /usr/share/emacs22/site-lisp/emacs-goodies-el/filladapt.el

  relpathshow /usr/share/emacs23/site-lisp/emacs-goodies-el \
              /usr/share/emacs22/site-lisp/emacs-goodies-el/filladapt.el

  relpathshow /usr/bin \
              /usr/share/emacs22/site-lisp/emacs-goodies-el/filladapt.el

  relpathshow /usr/bin \
              /usr/share/emacs22/site-lisp/emacs-goodies-el/filladapt.el

  relpathshow /usr/bin/share/emacs22/site-lisp/emacs-goodies-el \
              /etc/motd

  relpathshow / \
              /initrd.img
fi

#!/bin/bash
# both $1 and $2 are absolute paths
# returns $2 relative to $1

source=$1
target=$2

common_part=$source
back=
while [ "${target#$common_part}" = "${target}" ]; do
  common_part=$(dirname $common_part)
  back="../${back}"
done

echo ${back}${target#$common_part/}

这个答案并没有解决问题的Bash部分，但是因为我试图使用这个问题中的答案在Emacs中实现这个功能，所以我就把它扔到那里。

Emacs实际上有一个开箱即用的函数:

ELISP> (file-relative-name "/a/b/c" "/a/b/c")
"."
ELISP> (file-relative-name "/a/b/c" "/a/b")
"c"
ELISP> (file-relative-name "/a/b/c" "/c/b")
"../../a/b/c"