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
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...
其他回答
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。
我知道这是一个古老的问题。我仍在提供另一种选择。最近我遇到了同样的问题,并且发现没有现有的可移植命令来执行此操作。因此,我编写了下面的shell脚本,其中包括一个可以实现此功能的函数。
#! /bin/sh
function normalize {
local rc=0
local ret
if [ $# -gt 0 ] ; then
# invalid
if [ "x`echo $1 | grep -E '^/\.\.'`" != "x" ] ; then
echo $1
return -1
fi
# convert to absolute path
if [ "x`echo $1 | grep -E '^\/'`" == "x" ] ; then
normalize "`pwd`/$1"
return $?
fi
ret=`echo $1 | sed 's;/\.\($\|/\);/;g' | sed 's;/[^/]*[^/.]\+[^/]*/\.\.\($\|/\);/;g'`
else
read line
normalize "$line"
return $?
fi
if [ "x`echo $ret | grep -E '/\.\.?(/|$)'`" != "x" ] ; then
ret=`normalize "$ret"`
rc=$?
fi
echo "$ret"
return $rc
}
https://gist.github.com/bestofsong/8830bdf3e5eb9461d27313c3c282868c
我最近的解决方案是:
pushd foo/bar/..
dir=`pwd`
popd
根据蒂姆·惠特科姆的回答。
一个使用node.js的简单解决方案:
#!/usr/bin/env node
process.stdout.write(require('path').resolve(process.argv[2]));
正如Adam Liss所指出的,realpath并非与每个发行版都捆绑在一起。这很遗憾,因为这是最好的解决方案。提供的源代码很棒,我可能现在就开始使用它。以下是我到目前为止一直在使用的,只是为了完整起见,我在这里分享:
get_abs_path() {
local PARENT_DIR=$(dirname "$1")
cd "$PARENT_DIR"
local ABS_PATH="$(pwd)"/"$(basename "$1")"
cd - >/dev/null
echo "$ABS_PATH"
}
如果您希望它解析符号链接,只需将pwd替换为pwd -P。