Linux 你如何在 Bash 中规范化文件路径？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/284662/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How do you normalize a file path in Bash?
提问by Fabien
I want to transform /foo/bar/..to /foo
我想转变/foo/bar/..为/foo
Is there a bash command which does this?
是否有执行此操作的 bash 命令？
Edit: in my practical case, the directory does exist.
编辑：在我的实际情况下，该目录确实存在。
采纳答案by Kent Fredric
if you're wanting to chomp part of a filename from the path, "dirname" and "basename" are your friends, and "realpath" is handy too.
如果你想从路径中截取文件名的一部分，“dirname”和“basename”是你的朋友，“realpath”也很方便。
dirname /foo/bar/baz
# /foo/bar
basename /foo/bar/baz
# baz
dirname $( dirname /foo/bar/baz )
# /foo
realpath ../foo
# ../foo: No such file or directory
realpath /tmp/../tmp/../tmp
# /tmp
realpathalternatives
realpath备择方案
If realpathis not supported by your shell, you can try
如果realpath您的外壳不支持，您可以尝试
readlink -f /path/here/..
Also
还
readlink -m /path/there/../../
Works the same as
工作原理与
realpath -s /path/here/../../
in that the path doesn't need to exist to be normalized.
因为路径不需要存在即可标准化。
回答by Adam Liss
Try realpath. Below is the source in its entirety, hereby donated to the public domain.
试试realpath。以下是全部来源，特此捐赠给公共领域。
// realpath.c: display the absolute path to a file or directory.
// Adam Liss, August, 2007
// This program is provided "as-is" to the public domain, without express or
// implied warranty, for any non-profit use, provided this notice is maintained.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <libgen.h>
#include <limits.h>
static char *s_pMyName;
void usage(void);
int main(int argc, char *argv[])
{
char
sPath[PATH_MAX];
s_pMyName = strdup(basename(argv[0]));
if (argc < 2)
usage();
printf("%s\n", realpath(argv[1], sPath));
return 0;
}
void usage(void)
{
fprintf(stderr, "usage: %s PATH\n", s_pMyName);
exit(1);
}
回答by Tim Whitcomb
I don't know if there is a direct bash command to do this, but I usually do
我不知道是否有直接的 bash 命令来执行此操作，但我通常会这样做
normalDir="`cd "${dirToNormalize}";pwd`"
echo "${normalDir}"
and it works well.
它运作良好。
回答by alhernau
Talkative, and a bit late answer. I need to write one since I'm stuck on older RHEL4/5. I handles absolute and relative links, and simplifies //, /./ and somedir/../ entries.
健谈，回答有点晚。我需要写一个，因为我被困在旧的 RHEL4/5 上。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 | /bin/cut -d'>' -f 2)
}
test -x /usr/bin/realpath || realpath () {
local PATH=/bin:/usr/bin
local inputpath=
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 $realpath
}
mkdir -p /tmp/bar
(cd /tmp ; ln -s /tmp/bar foo; ln -s ../.././usr /tmp/bar/link2usr)
echo `realpath /tmp/foo`
回答by mattalxndr
Use the readlink utility from the coreutils package.
使用 coreutils 包中的 readlink 实用程序。
MY_PATH=$(readlink -f "get_abs_path() {
local PARENT_DIR=$(dirname "")
cd "$PARENT_DIR"
local ABS_PATH="$(pwd)"/"$(basename "")"
cd - >/dev/null
echo "$ABS_PATH"
}
")
回答by loevborg
A portable and reliable solution is to use python, which is preinstalled pretty much everywhere (including Darwin). You have two options:
一个可移植且可靠的解决方案是使用 python，它几乎在任何地方（包括 Darwin）都预装。您有两个选择：
abspathreturns an absolute path but does not resolve symlinks:python -c "import os,sys; print(os.path.abspath(sys.argv[1]))" path/to/filerealpathreturns an absolute path and in doing so resolves symlinks, generating a canonical path:python -c "import os,sys; print(os.path.realpath(sys.argv[1]))" path/to/file
abspath返回绝对路径但不解析符号链接：python -c "import os,sys; print(os.path.abspath(sys.argv[1]))" path/to/filerealpath返回一个绝对路径，并在这样做时解析符号链接，生成一个规范路径：python -c "import os,sys; print(os.path.realpath(sys.argv[1]))" path/to/file
In each case, path/to/filecan be either a relative or absolute path.
在每种情况下，path/to/file都可以是相对路径或绝对路径。
回答by Jeet
As Adam Liss noted realpathis not bundled with every distribution. Which is a shame, because it is the best solution. The provided source code is great, and I will probably start using it now. Here is what I have been using until now, which I share here just for completeness:
正如 Adam Liss 所指出的realpath，并不是每个发行版都捆绑在一起。这是一种耻辱，因为它是最好的解决方案。提供的源代码很棒，我现在可能会开始使用它。这是我到目前为止一直在使用的内容，为了完整起见，我在这里分享：
pushd foo/bar/..
dir=`pwd`
popd
If you want it to resolve symlinks, just replace pwdwith pwd -P.
如果您希望它解析符号链接，只需替换pwd为pwd -P.
回答by schmunk
My recent solution was:
我最近的解决方案是：
for f in $paths; do (cd $f; pwd); done
Based on the answer of Tim Whitcomb.
基于 Tim Whitcomb 的回答。
回答by Jesse Glick
Not exactly an answer but perhaps a follow-up question (original question was not explicit):
不完全是答案，但可能是后续问题（原始问题不明确）：
readlinkis fine if you actually want to follow symlinks. But there is also a use case for merely normalizing ./and ../and //sequences, which can be done purely syntactically, withoutcanonicalizing symlinks. readlinkis no good for this, and neither is realpath.
readlink如果您确实想遵循符号链接，那很好。但也有一个用例仅规范化./and../和//序列，这可以纯粹在语法上完成，而无需规范化符号链接。readlink对这没有好处，也不是realpath。
jrunscript -e 'for (var i = 0; i < arguments.length; i++) {println(new java.io.File(new java.io.File(arguments[i]).toURI().normalize()))}' $paths
works for existing paths, but breaks for others.
适用于现有路径，但适用于其他路径。
A sedscript would seem to be a good bet, except that you cannot iteratively replace sequences (/foo/bar/baz/../..-> /foo/bar/..-> /foo) without using something like Perl, which is not safe to assume on all systems, or using some ugly loop to compare the output of sedto its input.
一个sed脚本似乎是一个不错的选择，但你不能反复更换序列（/foo/bar/baz/../..- > /foo/bar/..- > /foo），而不使用像Perl中，这是不是安全地假定所有系统上，或者使用一些丑陋的环路的输出比较sed来它的输入。
FWIW, a one-liner using Java (JDK 6+):
FWIW，使用 Java (JDK 6+) 的单线：
resolve_dir() {
(builtin cd `dirname "${1/#~/$HOME}"`'/'`basename "${1/#~/$HOME}"` 2>/dev/null; if [ $? -eq 0 ]; then pwd; fi)
}
回答by apottere
I'm late to the party, but this is the solution I've crafted after reading a bunch of threads like this:
我迟到了，但这是我在阅读了一堆这样的线程后精心制作的解决方案：
##代码##This will resolve the absolute path of $1, play nice with ~, keep symlinks in the path where they are, and it won't mess with your directory stack. It returns the full path or nothing if it doesn't exist. It expects $1 to be a directory and will probably fail if it's not, but that's an easy check to do yourself.
这将解析 $1 的绝对路径，与 ~ 一起玩，将符号链接保留在它们所在的路径中，并且不会弄乱您的目录堆栈。如果不存在，则返回完整路径或不返回任何内容。它期望 $1 是一个目录，如果不是，则可能会失败，但您自己可以轻松检查。
