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
#!/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/}
其他回答
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
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"
Python的os.path.relpath作为shell函数
这个relpath练习的目标是模仿xni提出的Python 2.7的os.path.relpath函数(从Python 2.6版可用，但只能在2.7中正常工作)。因此，一些结果可能与其他答案中提供的函数不同。
(我没有在路径中测试换行符，因为它破坏了基于从ZSH调用python -c的验证。经过一些努力，这当然是可能的。)
关于Bash中的“魔法”，我很久以前就放弃了在Bash中寻找魔法，但我已经在ZSH中找到了我需要的所有魔法，然后是一些。
因此，我提出了两种实现。
第一个实现的目标是完全兼容posix。我已经在Debian 6.0.6的“挤压”上用/bin/dash测试了它。它还可以在OS X 10.8.3上完美地与/bin/sh一起工作，这实际上是伪装成POSIX shell的Bash版本3.2。
第二个实现是一个ZSH shell函数，它对路径中的多个斜杠和其他麻烦具有健壮性。如果您有可用的ZSH，这是推荐的版本，即使您是在下面给出的脚本形式中调用它(即使用#!/usr/bin/env zsh)。
最后，我编写了一个ZSH脚本，根据其他答案中提供的测试用例，验证$PATH中relpath命令的输出。我通过添加一些空格、制表符和标点符号，例如!？ *在这里和那里，还抛出了另一个测试与vim-powerline中发现的奇异UTF-8字符。
POSIX外壳函数
首先，posix兼容的shell函数。它适用于各种路径，但不清除多个斜杠或解析符号链接。
#!/bin/sh
relpath () {
[ $# -ge 1 ] && [ $# -le 2 ] || return 1
current="${2:+"$1"}"
target="${2:-"$1"}"
[ "$target" != . ] || target=/
target="/${target##/}"
[ "$current" != . ] || current=/
current="${current:="/"}"
current="/${current##/}"
appendix="${target##/}"
relative=''
while appendix="${target#"$current"/}"
[ "$current" != '/' ] && [ "$appendix" = "$target" ]; do
if [ "$current" = "$appendix" ]; then
relative="${relative:-.}"
echo "${relative#/}"
return 0
fi
current="${current%/*}"
relative="$relative${relative:+/}.."
done
relative="$relative${relative:+${appendix:+/}}${appendix#/}"
echo "$relative"
}
relpath "$@"
ZSH壳函数
现在是更加健壮的zsh版本。如果您希望它将参数解析为真实路径à la realpath -f(在Linux coreutils包中可用)，请将第3行和第4行上的:a替换为:a。
要在zsh中使用它，请删除第一行和最后一行，并将其放在$FPATH变量中的目录中。
#!/usr/bin/env zsh
relpath () {
[[ $# -ge 1 ]] && [[ $# -le 2 ]] || return 1
local target=${${2:-$1}:a} # replace `:a' by `:A` to resolve symlinks
local current=${${${2:+$1}:-$PWD}:a} # replace `:a' by `:A` to resolve symlinks
local appendix=${target#/}
local relative=''
while appendix=${target#$current/}
[[ $current != '/' ]] && [[ $appendix = $target ]]; do
if [[ $current = $appendix ]]; then
relative=${relative:-.}
print ${relative#/}
return 0
fi
current=${current%/*}
relative="$relative${relative:+/}.."
done
relative+=${relative:+${appendix:+/}}${appendix#/}
print $relative
}
relpath "$@"
测试脚本
最后是测试脚本。它接受一个选项，即-v来启用详细输出。
#!/usr/bin/env zsh
set -eu
VERBOSE=false
script_name=$(basename $0)
usage () {
print "\n Usage: $script_name SRC_PATH DESTINATION_PATH\n" >&2
exit ${1:=1}
}
vrb () { $VERBOSE && print -P ${(%)@} || return 0; }
relpath_check () {
[[ $# -ge 1 ]] && [[ $# -le 2 ]] || return 1
target=${${2:-$1}}
prefix=${${${2:+$1}:-$PWD}}
result=$(relpath $prefix $target)
# Compare with python's os.path.relpath function
py_result=$(python -c "import os.path; print os.path.relpath('$target', '$prefix')")
col='%F{green}'
if [[ $result != $py_result ]] && col='%F{red}' || $VERBOSE; then
print -P "${col}Source: '$prefix'\nDestination: '$target'%f"
print -P "${col}relpath: ${(qq)result}%f"
print -P "${col}python: ${(qq)py_result}%f\n"
fi
}
run_checks () {
print "Running checks..."
relpath_check '/ a b/å/⮀*/!' '/ a b/å/⮀/xäå/?'
relpath_check '/' '/A'
relpath_check '/A' '/'
relpath_check '/ & / !/*/\\/E' '/'
relpath_check '/' '/ & / !/*/\\/E'
relpath_check '/ & / !/*/\\/E' '/ & / !/?/\\/E/F'
relpath_check '/X/Y' '/ & / !/C/\\/E/F'
relpath_check '/ & / !/C' '/A'
relpath_check '/A / !/C' '/A /B'
relpath_check '/Â/ !/C' '/Â/ !/C'
relpath_check '/ & /B / C' '/ & /B / C/D'
relpath_check '/ & / !/C' '/ & / !/C/\\/Ê'
relpath_check '/Å/ !/C' '/Å/ !/D'
relpath_check '/.A /*B/C' '/.A /*B/\\/E'
relpath_check '/ & / !/C' '/ & /D'
relpath_check '/ & / !/C' '/ & /\\/E'
relpath_check '/ & / !/C' '/\\/E/F'
relpath_check /home/part1/part2 /home/part1/part3
relpath_check /home/part1/part2 /home/part4/part5
relpath_check /home/part1/part2 /work/part6/part7
relpath_check /home/part1 /work/part1/part2/part3/part4
relpath_check /home /work/part2/part3
relpath_check / /work/part2/part3/part4
relpath_check /home/part1/part2 /home/part1/part2/part3/part4
relpath_check /home/part1/part2 /home/part1/part2/part3
relpath_check /home/part1/part2 /home/part1/part2
relpath_check /home/part1/part2 /home/part1
relpath_check /home/part1/part2 /home
relpath_check /home/part1/part2 /
relpath_check /home/part1/part2 /work
relpath_check /home/part1/part2 /work/part1
relpath_check /home/part1/part2 /work/part1/part2
relpath_check /home/part1/part2 /work/part1/part2/part3
relpath_check /home/part1/part2 /work/part1/part2/part3/part4
relpath_check home/part1/part2 home/part1/part3
relpath_check home/part1/part2 home/part4/part5
relpath_check home/part1/part2 work/part6/part7
relpath_check home/part1 work/part1/part2/part3/part4
relpath_check home work/part2/part3
relpath_check . work/part2/part3
relpath_check home/part1/part2 home/part1/part2/part3/part4
relpath_check home/part1/part2 home/part1/part2/part3
relpath_check home/part1/part2 home/part1/part2
relpath_check home/part1/part2 home/part1
relpath_check home/part1/part2 home
relpath_check home/part1/part2 .
relpath_check home/part1/part2 work
relpath_check home/part1/part2 work/part1
relpath_check home/part1/part2 work/part1/part2
relpath_check home/part1/part2 work/part1/part2/part3
relpath_check home/part1/part2 work/part1/part2/part3/part4
print "Done with checks."
}
if [[ $# -gt 0 ]] && [[ $1 = "-v" ]]; then
VERBOSE=true
shift
fi
if [[ $# -eq 0 ]]; then
run_checks
else
VERBOSE=true
relpath_check "$@"
fi
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[$i]}"
fi
done
#One last newline
echo
另一个解决方案，纯bash + GNU readlink，在以下上下文中易于使用:
ln -s "$(relpath "$A" "$B")" "$B"
编辑:确保“$B”是不存在或没有软链接在这种情况下，否则relpath遵循这个链接，这不是你想要的!
这几乎适用于当前所有的Linux。如果readlink -m在您这边不起作用，请尝试readlink -f。请参见https://gist.github.com/hilbix/1ec361d00a8178ae8ea0查看可能的更新:
: relpath A B
# Calculate relative path from A to B, returns true on success
# Example: ln -s "$(relpath "$A" "$B")" "$B"
relpath()
{
local X Y A
# We can create dangling softlinks
X="$(readlink -m -- "$1")" || return
Y="$(readlink -m -- "$2")" || return
X="${X%/}/"
A=""
while Y="${Y%/*}"
[ ".${X#"$Y"/}" = ".$X" ]
do
A="../$A"
done
X="$A${X#"$Y"/}"
X="${X%/}"
echo "${X:-.}"
}
注:
Care was taken that it is safe against unwanted shell meta character expansion, in case filenames contain * or ?. The output is meant to be usable as the first argument to ln -s: relpath / / gives . and not the empty string relpath a a gives a, even if a happens to be a directory Most common cases were tested to give reasonable results, too. This solution uses string prefix matching, hence readlink is required to canonicalize paths. Thanks to readlink -m it works for not yet existing paths, too.
在旧系统上，readlink -m不可用，如果文件不存在，readlink -f将失败。所以你可能需要一些像这样的解决方法(未经测试!):
readlink_missing()
{
readlink -m -- "$1" && return
readlink -f -- "$1" && return
[ -e . ] && echo "$(readlink_missing "$(dirname "$1")")/$(basename "$1")"
}
这在$1包含的情况下是不正确的。或. .对于不存在的路径(如/doesnotexist/./a)，但它应该涵盖大多数情况。
(用readlink_missing替换上面的readlink -m——)
编辑，因为下面是反对票
下面是一个测试，这个函数确实是正确的:
check()
{
res="$(relpath "$2" "$1")"
[ ".$res" = ".$3" ] && return
printf ':WRONG: %-10q %-10q gives %q\nCORRECT %-10q %-10q gives %q\n' "$1" "$2" "$res" "$@"
}
# TARGET SOURCE RESULT
check "/A/B/C" "/A" ".."
check "/A/B/C" "/A.x" "../../A.x"
check "/A/B/C" "/A/B" "."
check "/A/B/C" "/A/B/C" "C"
check "/A/B/C" "/A/B/C/D" "C/D"
check "/A/B/C" "/A/B/C/D/E" "C/D/E"
check "/A/B/C" "/A/B/D" "D"
check "/A/B/C" "/A/B/D/E" "D/E"
check "/A/B/C" "/A/D" "../D"
check "/A/B/C" "/A/D/E" "../D/E"
check "/A/B/C" "/D/E/F" "../../D/E/F"
check "/foo/baz/moo" "/foo/bar" "../bar"
困惑吗?好吧，这是正确的结果!即使你认为它不符合问题，以下是正确的证明:
check "http://example.com/foo/baz/moo" "http://example.com/foo/bar" "../bar"
毫无疑问，……/bar是从页面moo中看到的页面栏的准确且唯一正确的相对路径。其他一切都是完全错误的。
采用问题的输出很简单，显然假设current是一个目录:
absolute="/foo/bar"
current="/foo/baz/foo"
relative="../$(relpath "$absolute" "$current")"
这将返回所请求的内容。
在你感到惊讶之前，这里有一个稍微复杂一点的relpath变体(注意细微的区别)，它也应该适用于url语法(因此，由于一些bash魔法，末尾/幸存下来):
# Calculate relative PATH to the given DEST from the given BASE
# In the URL case, both URLs must be absolute and have the same Scheme.
# The `SCHEME:` must not be present in the FS either.
# This way this routine works for file paths an
: relpathurl DEST BASE
relpathurl()
{
local X Y A
# We can create dangling softlinks
X="$(readlink -m -- "$1")" || return
Y="$(readlink -m -- "$2")" || return
X="${X%/}/${1#"${1%/}"}"
Y="${Y%/}${2#"${2%/}"}"
A=""
while Y="${Y%/*}"
[ ".${X#"$Y"/}" = ".$X" ]
do
A="../$A"
done
X="$A${X#"$Y"/}"
X="${X%/}"
echo "${X:-.}"
}
这里有一些检查，只是为了弄清楚:它确实像所说的那样工作。
check()
{
res="$(relpathurl "$2" "$1")"
[ ".$res" = ".$3" ] && return
printf ':WRONG: %-10q %-10q gives %q\nCORRECT %-10q %-10q gives %q\n' "$1" "$2" "$res" "$@"
}
# TARGET SOURCE RESULT
check "/A/B/C" "/A" ".."
check "/A/B/C" "/A.x" "../../A.x"
check "/A/B/C" "/A/B" "."
check "/A/B/C" "/A/B/C" "C"
check "/A/B/C" "/A/B/C/D" "C/D"
check "/A/B/C" "/A/B/C/D/E" "C/D/E"
check "/A/B/C" "/A/B/D" "D"
check "/A/B/C" "/A/B/D/E" "D/E"
check "/A/B/C" "/A/D" "../D"
check "/A/B/C" "/A/D/E" "../D/E"
check "/A/B/C" "/D/E/F" "../../D/E/F"
check "/foo/baz/moo" "/foo/bar" "../bar"
check "http://example.com/foo/baz/moo" "http://example.com/foo/bar" "../bar"
check "http://example.com/foo/baz/moo/" "http://example.com/foo/bar" "../../bar"
check "http://example.com/foo/baz/moo" "http://example.com/foo/bar/" "../bar/"
check "http://example.com/foo/baz/moo/" "http://example.com/foo/bar/" "../../bar/"
下面是如何用它从问题中得到想要的结果:
absolute="/foo/bar"
current="/foo/baz/foo"
relative="$(relpathurl "$absolute" "$current/")"
echo "$relative"
如果你发现什么东西不起作用，请在下面的评论中告诉我。谢谢。
PS:
为什么relpath的论点与这里的所有其他答案相反?
如果你改变
Y="$(readlink -m -- "$2")" || return
to
Y="$(readlink -m -- "${2:-"$PWD"}")" || return
然后你可以去掉第二个参数，这样BASE就是当前目录/URL/任何东西。这只是Unix的原则。
