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
我的解决方案:
computeRelativePath()
{
Source=$(readlink -f ${1})
Target=$(readlink -f ${2})
local OLDIFS=$IFS
IFS="/"
local SourceDirectoryArray=($Source)
local TargetDirectoryArray=($Target)
local SourceArrayLength=$(echo ${SourceDirectoryArray[@]} | wc -w)
local TargetArrayLength=$(echo ${TargetDirectoryArray[@]} | wc -w)
local Length
test $SourceArrayLength -gt $TargetArrayLength && Length=$SourceArrayLength || Length=$TargetArrayLength
local Result=""
local AppendToEnd=""
IFS=$OLDIFS
local i
for ((i = 0; i <= $Length + 1 ; i++ ))
do
if [ "${SourceDirectoryArray[$i]}" = "${TargetDirectoryArray[$i]}" ]
then
continue
elif [ "${SourceDirectoryArray[$i]}" != "" ] && [ "${TargetDirectoryArray[$i]}" != "" ]
then
AppendToEnd="${AppendToEnd}${TargetDirectoryArray[${i}]}/"
Result="${Result}../"
elif [ "${SourceDirectoryArray[$i]}" = "" ]
then
Result="${Result}${TargetDirectoryArray[${i}]}/"
else
Result="${Result}../"
fi
done
Result="${Result}${AppendToEnd}"
echo $Result
}
其他回答
这个脚本只对路径名有效。它不需要任何文件存在。如果传递的路径不是绝对的，那么行为就有点不寻常，但是如果两条路径都是相对的，那么应该能正常工作。
我只在OS X上测试过，所以可能不太便携。
#!/bin/bash
set -e
declare SCRIPT_NAME="$(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[$i]}"
fi
done
#One last newline
echo
下面是一个shell脚本，它可以在不调用其他程序的情况下完成:
#! /bin/env bash
#bash script to find the relative path between two directories
mydir=${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
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#$s/}
我从pini和其他一些想法中得到了答案
注意:这要求两个路径都是现有文件夹。文件将无法工作。
另一个解决方案，纯bash + GNU readlink，在以下上下文中易于使用:
ln -s "$(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。
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。
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