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
该脚本仅对绝对路径或没有绝对路径的相对路径的输入提供正确的结果。或者. .:
#!/bin/bash
# usage: relpath from to
if [[ "$1" == "$2" ]]
then
echo "."
exit
fi
IFS="/"
current=($1)
absolute=($2)
abssize=${#absolute[@]}
cursize=${#current[@]}
while [[ ${absolute[level]} == ${current[level]} ]]
do
(( level++ ))
if (( level > abssize || level > cursize ))
then
break
fi
done
for ((i = level; i < cursize; i++))
do
if ((i > level))
then
newpath=$newpath"/"
fi
newpath=$newpath".."
done
for ((i = level; i < abssize; i++))
do
if [[ -n $newpath ]]
then
newpath=$newpath"/"
fi
newpath=$newpath${absolute[i]}
done
echo "$newpath"
其他回答
这个答案并没有解决问题的Bash部分，但是因为我试图使用这个问题中的答案在Emacs中实现这个功能，所以我就把它扔到那里。
Emacs实际上有一个开箱即用的函数:
ELISP> (file-relative-name "/a/b/c" "/a/b/c")
"."
ELISP> (file-relative-name "/a/b/c" "/a/b")
"c"
ELISP> (file-relative-name "/a/b/c" "/c/b")
"../../a/b/c"
test.sh:
#!/bin/bash
cd /home/ubuntu
touch blah
TEST=/home/ubuntu/.//blah
echo TEST=$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
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 :
可悲的是，Mark Rushakoff的答案(现在删除了-它引用了这里的代码)在适应时似乎不能正确工作:
source=/home/part2/part3/part4
target=/work/proj1/proj2
注释中概述的思想可以经过改进，使其在大多数情况下正确工作。我假设脚本有一个源参数(您所在的位置)和一个目标参数(您想要到达的位置)，这两个参数要么是绝对路径名，要么是相对路径名。如果一个是绝对的，另一个是相对的，最简单的方法是用当前工作目录作为相对名称的前缀——但是下面的代码没有这样做。
当心
下面的代码接近正确工作，但不是很正确。
丹尼斯·威廉姆森(Dennis Williamson)的评论中提到了一个问题。 还有一个问题，这种纯文本的路径名处理，你可能会被奇怪的符号链接搞得一团糟。 该代码不处理'xyz/./pqr'等路径中的零散'点'。 该代码不处理'xyz/../pqr'等路径中的流浪'双点'。 简单地说:代码没有删除前导'。/'从路径。
Dennis的代码更好，因为它修复了1和5，但也有相同的问题2,3,4。 因此，请使用Dennis的代码(并在此之前对其进行投票)。
注意:POSIX提供了一个系统调用realpath()来解析路径名，这样路径名中就没有符号链接了。将其应用于输入名称，然后使用Dennis的代码每次都会给出正确的答案。编写包装realpath()的C代码很简单——我已经做过了——但我不知道有什么标准实用程序可以这样做。)
为此，我发现Perl比shell更容易使用，尽管bash对数组有很好的支持，并且可能也可以做到这一点——这是读者的练习。因此，给定两个兼容的名称，将它们分别分解为组件:
Set the relative path to empty. While the components are the same, skip to the next. When corresponding components are different or there are no more components for one path: If there are no remaining source components and the relative path is empty, add "." to the start. For each remaining source component, prefix the relative path with "../". If there are no remaining target components and the relative path is empty, add "." to the start. For each remaining target component, add the component to the end of the path after a slash.
因此:
#!/bin/perl -w
use strict;
# Should fettle the arguments if one is absolute and one relative:
# Oops - missing functionality!
# Split!
my(@source) = split '/', $ARGV[0];
my(@target) = split '/', $ARGV[1];
my $count = scalar(@source);
$count = scalar(@target) if (scalar(@target) < $count);
my $relpath = "";
my $i;
for ($i = 0; $i < $count; $i++)
{
last if $source[$i] ne $target[$i];
}
$relpath = "." if ($i >= scalar(@source) && $relpath eq "");
for (my $s = $i; $s < scalar(@source); $s++)
{
$relpath = "../$relpath";
}
$relpath = "." if ($i >= scalar(@target) && $relpath eq "");
for (my $t = $i; $t < scalar(@target); $t++)
{
$relpath .= "/$target[$t]";
}
# Clean up result (remove double slash, trailing slash, trailing slash-dot).
$relpath =~ s%//%/%;
$relpath =~ s%/$%%;
$relpath =~ s%/\.$%%;
print "source = $ARGV[0]\n";
print "target = $ARGV[1]\n";
print "relpath = $relpath\n";
测试脚本(方括号包含一个空格和一个制表符):
sed 's/#.*//;/^[ ]*$/d' <<! |
/home/part1/part2 /home/part1/part3
/home/part1/part2 /home/part4/part5
/home/part1/part2 /work/part6/part7
/home/part1 /work/part1/part2/part3/part4
/home /work/part2/part3
/ /work/part2/part3/part4
/home/part1/part2 /home/part1/part2/part3/part4
/home/part1/part2 /home/part1/part2/part3
/home/part1/part2 /home/part1/part2
/home/part1/part2 /home/part1
/home/part1/part2 /home
/home/part1/part2 /
/home/part1/part2 /work
/home/part1/part2 /work/part1
/home/part1/part2 /work/part1/part2
/home/part1/part2 /work/part1/part2/part3
/home/part1/part2 /work/part1/part2/part3/part4
home/part1/part2 home/part1/part3
home/part1/part2 home/part4/part5
home/part1/part2 work/part6/part7
home/part1 work/part1/part2/part3/part4
home work/part2/part3
. work/part2/part3
home/part1/part2 home/part1/part2/part3/part4
home/part1/part2 home/part1/part2/part3
home/part1/part2 home/part1/part2
home/part1/part2 home/part1
home/part1/part2 home
home/part1/part2 .
home/part1/part2 work
home/part1/part2 work/part1
home/part1/part2 work/part1/part2
home/part1/part2 work/part1/part2/part3
home/part1/part2 work/part1/part2/part3/part4
!
while read source target
do
perl relpath.pl $source $target
echo
done
测试脚本的输出:
source = /home/part1/part2
target = /home/part1/part3
relpath = ../part3
source = /home/part1/part2
target = /home/part4/part5
relpath = ../../part4/part5
source = /home/part1/part2
target = /work/part6/part7
relpath = ../../../work/part6/part7
source = /home/part1
target = /work/part1/part2/part3/part4
relpath = ../../work/part1/part2/part3/part4
source = /home
target = /work/part2/part3
relpath = ../work/part2/part3
source = /
target = /work/part2/part3/part4
relpath = ./work/part2/part3/part4
source = /home/part1/part2
target = /home/part1/part2/part3/part4
relpath = ./part3/part4
source = /home/part1/part2
target = /home/part1/part2/part3
relpath = ./part3
source = /home/part1/part2
target = /home/part1/part2
relpath = .
source = /home/part1/part2
target = /home/part1
relpath = ..
source = /home/part1/part2
target = /home
relpath = ../..
source = /home/part1/part2
target = /
relpath = ../../../..
source = /home/part1/part2
target = /work
relpath = ../../../work
source = /home/part1/part2
target = /work/part1
relpath = ../../../work/part1
source = /home/part1/part2
target = /work/part1/part2
relpath = ../../../work/part1/part2
source = /home/part1/part2
target = /work/part1/part2/part3
relpath = ../../../work/part1/part2/part3
source = /home/part1/part2
target = /work/part1/part2/part3/part4
relpath = ../../../work/part1/part2/part3/part4
source = home/part1/part2
target = home/part1/part3
relpath = ../part3
source = home/part1/part2
target = home/part4/part5
relpath = ../../part4/part5
source = home/part1/part2
target = work/part6/part7
relpath = ../../../work/part6/part7
source = home/part1
target = work/part1/part2/part3/part4
relpath = ../../work/part1/part2/part3/part4
source = home
target = work/part2/part3
relpath = ../work/part2/part3
source = .
target = work/part2/part3
relpath = ../work/part2/part3
source = home/part1/part2
target = home/part1/part2/part3/part4
relpath = ./part3/part4
source = home/part1/part2
target = home/part1/part2/part3
relpath = ./part3
source = home/part1/part2
target = home/part1/part2
relpath = .
source = home/part1/part2
target = home/part1
relpath = ..
source = home/part1/part2
target = home
relpath = ../..
source = home/part1/part2
target = .
relpath = ../../..
source = home/part1/part2
target = work
relpath = ../../../work
source = home/part1/part2
target = work/part1
relpath = ../../../work/part1
source = home/part1/part2
target = work/part1/part2
relpath = ../../../work/part1/part2
source = home/part1/part2
target = work/part1/part2/part3
relpath = ../../../work/part1/part2/part3
source = home/part1/part2
target = work/part1/part2/part3/part4
relpath = ../../../work/part1/part2/part3/part4
面对奇怪的输入，这个Perl脚本在Unix上运行得相当彻底(它没有考虑Windows路径名的所有复杂性)。它使用模块Cwd及其函数realpath解析存在的名称的真实路径，并对不存在的路径进行文本分析。在所有情况下，除了一种情况，它产生的输出都与Dennis的脚本相同。越轨的情况是:
source = home/part1/part2
target = .
relpath1 = ../../..
relpath2 = ../../../.
这两个结果是等价的，只是不完全相同。(输出来自测试脚本的一个轻微修改版本——下面的Perl脚本只是输出答案，而不是像上面的脚本那样输出输入和答案。)现在，我应该排除无效的答案吗?也许……
#!/bin/perl -w
# Based loosely on code from: http://unix.derkeiler.com/Newsgroups/comp.unix.shell/2005-10/1256.html
# Via: http://stackoverflow.com/questions/2564634
use strict;
die "Usage: $0 from to\n" if scalar @ARGV != 2;
use Cwd qw(realpath getcwd);
my $pwd;
my $verbose = 0;
# Fettle filename so it is absolute.
# Deals with '//', '/./' and '/../' notations, plus symlinks.
# The realpath() function does the hard work if the path exists.
# For non-existent paths, the code does a purely textual hack.
sub resolve
{
my($name) = @_;
my($path) = realpath($name);
if (!defined $path)
{
# Path does not exist - do the best we can with lexical analysis
# Assume Unix - not dealing with Windows.
$path = $name;
if ($name !~ m%^/%)
{
$pwd = getcwd if !defined $pwd;
$path = "$pwd/$path";
}
$path =~ s%//+%/%g; # Not UNC paths.
$path =~ s%/$%%; # No trailing /
$path =~ s%/\./%/%g; # No embedded /./
# Try to eliminate /../abc/
$path =~ s%/\.\./(?:[^/]+)(/|$)%$1%g;
$path =~ s%/\.$%%; # No trailing /.
$path =~ s%^\./%%; # No leading ./
# What happens with . and / as inputs?
}
return($path);
}
sub print_result
{
my($source, $target, $relpath) = @_;
if ($verbose)
{
print "source = $ARGV[0]\n";
print "target = $ARGV[1]\n";
print "relpath = $relpath\n";
}
else
{
print "$relpath\n";
}
exit 0;
}
my($source) = resolve($ARGV[0]);
my($target) = resolve($ARGV[1]);
print_result($source, $target, ".") if ($source eq $target);
# Split!
my(@source) = split '/', $source;
my(@target) = split '/', $target;
my $count = scalar(@source);
$count = scalar(@target) if (scalar(@target) < $count);
my $relpath = "";
my $i;
# Both paths are absolute; Perl splits an empty field 0.
for ($i = 1; $i < $count; $i++)
{
last if $source[$i] ne $target[$i];
}
for (my $s = $i; $s < scalar(@source); $s++)
{
$relpath = "$relpath/" if ($s > $i);
$relpath = "$relpath..";
}
for (my $t = $i; $t < scalar(@target); $t++)
{
$relpath = "$relpath/" if ($relpath ne "");
$relpath = "$relpath$target[$t]";
}
print_result($source, $target, $relpath);
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"
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