我想转换/foo/bar/../ foo
是否有bash命令可以做到这一点?
编辑:在我的实际案例中,目录确实存在。
我想转换/foo/bar/../ foo
是否有bash命令可以做到这一点?
编辑:在我的实际案例中,目录确实存在。
当前回答
一个使用node.js的简单解决方案:
#!/usr/bin/env node
process.stdout.write(require('path').resolve(process.argv[2]));
其他回答
话多,回答有点晚。我需要写一个,因为我卡住了旧的RHEL4/5。 I处理绝对和相对链接,并简化//,/。/和somedir/../条目。
test -x /usr/bin/readlink || readlink () {
echo $(/bin/ls -l $1 | /bin/cut -d'>' -f 2)
}
test -x /usr/bin/realpath || realpath () {
local PATH=/bin:/usr/bin
local inputpath=$1
local changemade=1
while [ $changemade -ne 0 ]
do
changemade=0
local realpath=""
local token=
for token in ${inputpath//\// }
do
case $token in
""|".") # noop
;;
"..") # up one directory
changemade=1
realpath=$(dirname $realpath)
;;
*)
if [ -h $realpath/$token ]
then
changemade=1
target=`readlink $realpath/$token`
if [ "${target:0:1}" = '/' ]
then
realpath=$target
else
realpath="$realpath/$target"
fi
else
realpath="$realpath/$token"
fi
;;
esac
done
inputpath=$realpath
done
echo $realpath
}
mkdir -p /tmp/bar
(cd /tmp ; ln -s /tmp/bar foo; ln -s ../.././usr /tmp/bar/link2usr)
echo `realpath /tmp/foo`
基于@Andre的回答,我可能有一个稍微更好的版本,以防有人在一个无循环,完全基于字符串操作的解决方案。对于那些不想取消对任何符号链接的引用(这是使用realpath或readlink -f的缺点)的人来说,它也很有用。
它适用于bash 3.2.25及更高版本。
shopt -s extglob
normalise_path() {
local path="$1"
# get rid of /../ example: /one/../two to /two
path="${path//\/*([!\/])\/\.\./}"
# get rid of /./ and //* example: /one/.///two to /one/two
path="${path//@(\/\.\/|\/+(\/))//}"
# remove the last '/.'
echo "${path%%/.}"
}
$ normalise_path /home/codemedic/../codemedic////.config
/home/codemedic/.config
如果你只想规范化一个路径,不管是否存在,不涉及文件系统,不解析任何链接,也不使用外部utils,这里有一个从Python的posixpath.normpath转换而来的纯Bash函数。
#!/usr/bin/env bash
# Normalize path, eliminating double slashes, etc.
# Usage: new_path="$(normpath "${old_path}")"
# Translated from Python's posixpath.normpath:
# https://github.com/python/cpython/blob/master/Lib/posixpath.py#L337
normpath() {
local IFS=/ initial_slashes='' comp comps=()
if [[ $1 == /* ]]; then
initial_slashes='/'
[[ $1 == //* && $1 != ///* ]] && initial_slashes='//'
fi
for comp in $1; do
[[ -z ${comp} || ${comp} == '.' ]] && continue
if [[ ${comp} != '..' || (-z ${initial_slashes} && ${#comps[@]} -eq 0) || (\
${#comps[@]} -gt 0 && ${comps[-1]} == '..') ]]; then
comps+=("${comp}")
elif ((${#comps[@]})); then
unset 'comps[-1]'
fi
done
comp="${initial_slashes}${comps[*]}"
printf '%s\n' "${comp:-.}"
}
例子:
new_path="$(normpath '/foo/bar/..')"
echo "${new_path}"
# /foo
normpath "relative/path/with trailing slashs////"
# relative/path/with trailing slashs
normpath "////a/../lot/././/mess////./here/./../"
# /lot/mess
normpath ""
# .
# (empty path resolved to dot)
Personally, I cannot understand why Shell, a language often used for manipulating files, doesn't offer basic functions to deal with paths. In python, we have nice libraries like os.path or pathlib, which offers a whole bunch of tools to extract filename, extension, basename, path segments, split or join paths, to get absolute or normalized paths, to determine relations between paths, to do everything without much brain. And they take care of edge cases, and they're reliable. In Shell, to do any of these, either we call external executables, or we have to reinvent wheels with these extremely rudimentary and arcane syntaxes...
FILEPATH="file.txt"
echo $(realpath $(dirname $FILEPATH))/$(basename $FILEPATH)
即使文件不存在,这也可以工作。它需要包含该文件的目录存在。
如果您想要从路径中删除部分文件名,“dirname”和“basename”是您的朋友,“realpath”也很方便。
dirname /foo/bar/baz
# /foo/bar
basename /foo/bar/baz
# baz
dirname $( dirname /foo/bar/baz )
# /foo
realpath ../foo
# ../foo: No such file or directory
realpath /tmp/../tmp/../tmp
# /tmp
realpath替代品
如果您的shell不支持realpath,您可以尝试
readlink -f /path/here/..
Also
readlink -m /path/there/../../
工作原理与
realpath -s /path/here/../../
在这种情况下,路径不需要存在就可以被规范化。