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
}

test.sh:

#!/bin/bash                                                                 

cd /home/ubuntu
touch blah
TEST=/home/ubuntu/.//blah
echo TEST=$TEST
TMP=$(readlink -e "$TEST")
echo TMP=$TMP
REL=${TMP#$(pwd)/}
echo REL=$REL

测试:

$ ./test.sh 
TEST=/home/ubuntu/.//blah
TMP=/home/ubuntu/blah
REL=blah

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。

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