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…)?
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
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
其他回答
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。
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"
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"
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。