例子:

absolute="/foo/bar"
current="/foo/baz/foo"

# Magic

relative="../../bar"

我如何创造魔法(希望不是太复杂的代码…)?


当前回答

kasku和Pini的答案略有改进,空格更好,允许传递相对路径:

#!/bin/bash
# both $1 and $2 are paths
# returns $2 relative to $1
absolute=`readlink -f "$2"`
current=`readlink -f "$1"`
# Perl is magic
# Quoting horror.... spaces cause problems, that's why we need the extra " in here:
relative=$(perl -MFile::Spec -e "print File::Spec->abs2rel(q($absolute),q($current))")

echo $relative

其他回答

下面是一个shell脚本,它可以在不调用其他程序的情况下完成:

#! /bin/env bash 

#bash script to find the relative path between two directories

mydir=${0%/}
mydir=${0%/*}
creadlink="$mydir/creadlink"

shopt -s extglob

relpath_ () {
        path1=$("$creadlink" "$1")
        path2=$("$creadlink" "$2")
        orig1=$path1
        path1=${path1%/}/
        path2=${path2%/}/

        while :; do
                if test ! "$path1"; then
                        break
                fi
                part1=${path2#$path1}
                if test "${part1#/}" = "$part1"; then
                        path1=${path1%/*}
                        continue
                fi
                if test "${path2#$path1}" = "$path2"; then
                        path1=${path1%/*}
                        continue
                fi
                break
        done
        part1=$path1
        path1=${orig1#$part1}
        depth=${path1//+([^\/])/..}
        path1=${path2#$path1}
        path1=${depth}${path2#$part1}
        path1=${path1##+(\/)}
        path1=${path1%/}
        if test ! "$path1"; then
                path1=.
        fi
        printf "$path1"

}

relpath_test () {
        res=$(relpath_ /path1/to/dir1 /path1/to/dir2 )
        expected='../dir2'
        test_results "$res" "$expected"

        res=$(relpath_ / /path1/to/dir2 )
        expected='path1/to/dir2'
        test_results "$res" "$expected"

        res=$(relpath_ /path1/to/dir2 / )
        expected='../../..'
        test_results "$res" "$expected"

        res=$(relpath_ / / )
        expected='.'
        test_results "$res" "$expected"

        res=$(relpath_ /path/to/dir2/dir3 /path/to/dir1/dir4/dir4a )
        expected='../../dir1/dir4/dir4a'
        test_results "$res" "$expected"

        res=$(relpath_ /path/to/dir1/dir4/dir4a /path/to/dir2/dir3 )
        expected='../../../dir2/dir3'
        test_results "$res" "$expected"

        #res=$(relpath_ . /path/to/dir2/dir3 )
        #expected='../../../dir2/dir3'
        #test_results "$res" "$expected"
}

test_results () {
        if test ! "$1" = "$2"; then
                printf 'failed!\nresult:\nX%sX\nexpected:\nX%sX\n\n' "$@"
        fi
}

#relpath_test

来源:http://www.ynform.org/w/Pub/Relpath

我将只使用Perl来完成这个不那么简单的任务:

absolute="/foo/bar"
current="/foo/baz/foo"

# Perl is magic
relative=$(perl -MFile::Spec -e 'print File::Spec->abs2rel("'$absolute'","'$current'")')

这是对@pini目前评分最高的解决方案(遗憾的是,它只处理少数情况)的更正,全功能改进

提醒:'-z'测试如果字符串是零长度(=空),'-n'测试如果字符串不是空。

# both $1 and $2 are absolute paths beginning with /
# returns relative path to $2/$target from $1/$source
source=$1
target=$2

common_part=$source # for now
result="" # for now

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

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

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

# and now stick all parts together
if [[ -n $result ]] && [[ -n $forward_part ]]; then
    result="$result$forward_part"
elif [[ -n $forward_part ]]; then
    # extra slash removal
    result="${forward_part:1}"
fi

echo $result

测试用例:

compute_relative.sh "/A/B/C" "/A"           -->  "../.."
compute_relative.sh "/A/B/C" "/A/B"         -->  ".."
compute_relative.sh "/A/B/C" "/A/B/C"       -->  ""
compute_relative.sh "/A/B/C" "/A/B/C/D"     -->  "D"
compute_relative.sh "/A/B/C" "/A/B/C/D/E"   -->  "D/E"
compute_relative.sh "/A/B/C" "/A/B/D"       -->  "../D"
compute_relative.sh "/A/B/C" "/A/B/D/E"     -->  "../D/E"
compute_relative.sh "/A/B/C" "/A/D"         -->  "../../D"
compute_relative.sh "/A/B/C" "/A/D/E"       -->  "../../D/E"
compute_relative.sh "/A/B/C" "/D/E/F"       -->  "../../../D/E/F"

我需要这样的东西,但它也解决了符号链接。我发现pwd有一个-P标志用于此目的。附加了我的脚本的一个片段。它在shell脚本的函数中,因此是$1和$2。结果值是从START_ABS到END_ABS的相对路径,位于UPDIRS变量中。为了执行pwd -P,将脚本cd放入每个参数目录,这也意味着将处理相对路径参数。干杯,吉姆

SAVE_DIR="$PWD"
cd "$1"
START_ABS=`pwd -P`
cd "$SAVE_DIR"
cd "$2"
END_ABS=`pwd -P`

START_WORK="$START_ABS"
UPDIRS=""

while test -n "${START_WORK}" -a "${END_ABS/#${START_WORK}}" '==' "$END_ABS";
do
    START_WORK=`dirname "$START_WORK"`"/"
    UPDIRS=${UPDIRS}"../"
done
UPDIRS="$UPDIRS${END_ABS/#${START_WORK}}"
cd "$SAVE_DIR"

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