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

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。

其他回答

我把你的问题作为一个挑战,用“可移植的”shell代码来编写它,即。

考虑到POSIX外壳 没有数组之类的bashisms 避免像打瘟疫一样打外部电话。脚本中没有一个分叉!这使得它非常快,特别是在有显著分叉开销的系统上,比如cygwin。 必须处理路径名中的glob字符(*,?,[,])

它运行在任何POSIX兼容shell (zsh, bash, ksh, ash, busybox,…)上。它甚至包含一个测试套件来验证其操作。路径名的规范化留作练习。: -)

#!/bin/sh

# Find common parent directory path for a pair of paths.
# Call with two pathnames as args, e.g.
# commondirpart foo/bar foo/baz/bat -> result="foo/"
# The result is either empty or ends with "/".
commondirpart () {
   result=""
   while test ${#1} -gt 0 -a ${#2} -gt 0; do
      if test "${1%${1#?}}" != "${2%${2#?}}"; then   # First characters the same?
         break                                       # No, we're done comparing.
      fi
      result="$result${1%${1#?}}"                    # Yes, append to result.
      set -- "${1#?}" "${2#?}"                       # Chop first char off both strings.
   done
   case "$result" in
   (""|*/) ;;
   (*)     result="${result%/*}/";;
   esac
}

# Turn foo/bar/baz into ../../..
#
dir2dotdot () {
   OLDIFS="$IFS" IFS="/" result=""
   for dir in $1; do
      result="$result../"
   done
   result="${result%/}"
   IFS="$OLDIFS"
}

# Call with FROM TO args.
relativepath () {
   case "$1" in
   (*//*|*/./*|*/../*|*?/|*/.|*/..)
      printf '%s\n' "'$1' not canonical"; exit 1;;
   (/*)
      from="${1#?}";;
   (*)
      printf '%s\n' "'$1' not absolute"; exit 1;;
   esac
   case "$2" in
   (*//*|*/./*|*/../*|*?/|*/.|*/..)
      printf '%s\n' "'$2' not canonical"; exit 1;;
   (/*)
      to="${2#?}";;
   (*)
      printf '%s\n' "'$2' not absolute"; exit 1;;
   esac

   case "$to" in
   ("$from")   # Identical directories.
      result=".";;
   ("$from"/*) # From /x to /x/foo/bar -> foo/bar
      result="${to##$from/}";;
   ("")        # From /foo/bar to / -> ../..
      dir2dotdot "$from";;
   (*)
      case "$from" in
      ("$to"/*)       # From /x/foo/bar to /x -> ../..
         dir2dotdot "${from##$to/}";;
      (*)             # Everything else.
         commondirpart "$from" "$to"
         common="$result"
         dir2dotdot "${from#$common}"
         result="$result/${to#$common}"
      esac
      ;;
   esac
}

set -f # noglob

set -x
cat <<EOF |
/ / .
/- /- .
/? /? .
/?? /?? .
/??? /??? .
/?* /?* .
/* /* .
/* /** ../**
/* /*** ../***
/*.* /*.** ../*.**
/*.??? /*.?? ../*.??
/[] /[] .
/[a-z]* /[0-9]* ../[0-9]*
/foo /foo .
/foo / ..
/foo/bar / ../..
/foo/bar /foo ..
/foo/bar /foo/baz ../baz
/foo/bar /bar/foo  ../../bar/foo
/foo/bar/baz /gnarf/blurfl/blubb ../../../gnarf/blurfl/blubb
/foo/bar/baz /gnarf ../../../gnarf
/foo/bar/baz /foo/baz ../../baz
/foo. /bar. ../bar.
EOF
while read FROM TO VIA; do
   relativepath "$FROM" "$TO"
   printf '%s\n' "FROM: $FROM" "TO:   $TO" "VIA:  $result"
   if test "$result" != "$VIA"; then
      printf '%s\n' "OOOPS! Expected '$VIA' but got '$result'"
   fi
done

# vi: set tabstop=3 shiftwidth=3 expandtab fileformat=unix :

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

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。

$ 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