我如何找到一个Bash脚本位于该脚本内部的目录的路径?
我想用Bash脚本作为另一个应用程序的启动器,我想将工作目录更改为Bash脚本所在的目录,所以我可以在该目录中的文件上运行,如下:
$ ./application
我如何找到一个Bash脚本位于该脚本内部的目录的路径?
我想用Bash脚本作为另一个应用程序的启动器,我想将工作目录更改为Bash脚本所在的目录,所以我可以在该目录中的文件上运行,如下:
$ ./application
当前回答
如果不是由父母脚本来源,而不是同链接,0美元就足够了:
script_path="$0"
如果源于父母脚本而不是同链接,请使用 $BASH_SOURCE 或 ${BASH_SOURCE[0]}:
script_path="$BASH_SOURCE"
如果是同链接,请使用 $BASH_SOURCE 与 realpath 或 readlink -f 获取真正的文件路径:
script_path="$(realpath "$BASH_SOURCE")"
此外,路径或 readlink -f 返回绝对路径。
要获取脚本的目录,使用 dirname:
script_directory="$(dirname "$script_path")"
笔记
对于 MacOS 而言,请在这里或在这里找到一个替代路径或阅读链接 -f. 要使代码与不为 Bash 的支柱兼容,请使用 ${var-string} 参数扩展。
其他回答
我相信我得到了这个一个,我迟到了派对,但我认为有些人会欣赏它在这里,如果他们通过这个条纹。
#!/bin/sh # dash bash ksh # !zsh (issues). G. Nixon, 12/2013. Public domain.
## 'linkread' or 'fullpath' or (you choose) is a little tool to recursively
## dereference symbolic links (ala 'readlink') until the originating file
## is found. This is effectively the same function provided in stdlib.h as
## 'realpath' and on the command line in GNU 'readlink -f'.
## Neither of these tools, however, are particularly accessible on the many
## systems that do not have the GNU implementation of readlink, nor ship
## with a system compiler (not to mention the requisite knowledge of C).
## This script is written with portability and (to the extent possible, speed)
## in mind, hence the use of printf for echo and case statements where they
## can be substituded for test, though I've had to scale back a bit on that.
## It is (to the best of my knowledge) written in standard POSIX shell, and
## has been tested with bash-as-bin-sh, dash, and ksh93. zsh seems to have
## issues with it, though I'm not sure why; so probably best to avoid for now.
## Particularly useful (in fact, the reason I wrote this) is the fact that
## it can be used within a shell script to find the path of the script itself.
## (I am sure the shell knows this already; but most likely for the sake of
## security it is not made readily available. The implementation of "$0"
## specificies that the $0 must be the location of **last** symbolic link in
## a chain, or wherever it resides in the path.) This can be used for some
## ...interesting things, like self-duplicating and self-modifiying scripts.
## Currently supported are three errors: whether the file specified exists
## (ala ENOENT), whether its target exists/is accessible; and the special
## case of when a sybolic link references itself "foo -> foo": a common error
## for beginners, since 'ln' does not produce an error if the order of link
## and target are reversed on the command line. (See POSIX signal ELOOP.)
## It would probably be rather simple to write to use this as a basis for
## a pure shell implementation of the 'symlinks' util included with Linux.
## As an aside, the amount of code below **completely** belies the amount
## effort it took to get this right -- but I guess that's coding for you.
##===-------------------------------------------------------------------===##
for argv; do :; done # Last parameter on command line, for options parsing.
## Error messages. Use functions so that we can sub in when the error occurs.
recurses(){ printf "Self-referential:\n\t$argv ->\n\t$argv\n" ;}
dangling(){ printf "Broken symlink:\n\t$argv ->\n\t"$(readlink "$argv")"\n" ;}
errnoent(){ printf "No such file: "$@"\n" ;} # Borrow a horrible signal name.
# Probably best not to install as 'pathfull', if you can avoid it.
pathfull(){ cd "$(dirname "$@")"; link="$(readlink "$(basename "$@")")"
## 'test and 'ls' report different status for bad symlinks, so we use this.
if [ ! -e "$@" ]; then if $(ls -d "$@" 2>/dev/null) 2>/dev/null; then
errnoent 1>&2; exit 1; elif [ ! -e "$@" -a "$link" = "$@" ]; then
recurses 1>&2; exit 1; elif [ ! -e "$@" ] && [ ! -z "$link" ]; then
dangling 1>&2; exit 1; fi
fi
## Not a link, but there might be one in the path, so 'cd' and 'pwd'.
if [ -z "$link" ]; then if [ "$(dirname "$@" | cut -c1)" = '/' ]; then
printf "$@\n"; exit 0; else printf "$(pwd)/$(basename "$@")\n"; fi; exit 0
fi
## Walk the symlinks back to the origin. Calls itself recursivly as needed.
while [ "$link" ]; do
cd "$(dirname "$link")"; newlink="$(readlink "$(basename "$link")")"
case "$newlink" in
"$link") dangling 1>&2 && exit 1 ;;
'') printf "$(pwd)/$(basename "$link")\n"; exit 0 ;;
*) link="$newlink" && pathfull "$link" ;;
esac
done
printf "$(pwd)/$(basename "$newlink")\n"
}
## Demo. Install somewhere deep in the filesystem, then symlink somewhere
## else, symlink again (maybe with a different name) elsewhere, and link
## back into the directory you started in (or something.) The absolute path
## of the script will always be reported in the usage, along with "$0".
if [ -z "$argv" ]; then scriptname="$(pathfull "$0")"
# Yay ANSI l33t codes! Fancy.
printf "\n\033[3mfrom/as: \033[4m$0\033[0m\n\n\033[1mUSAGE:\033[0m "
printf "\033[4m$scriptname\033[24m [ link | file | dir ]\n\n "
printf "Recursive readlink for the authoritative file, symlink after "
printf "symlink.\n\n\n \033[4m$scriptname\033[24m\n\n "
printf " From within an invocation of a script, locate the script's "
printf "own file\n (no matter where it has been linked or "
printf "from where it is being called).\n\n"
else pathfull "$@"
fi
使用“0”字母:
#!/usr/bin/env bash
echo "The script you are running has basename $( basename -- "$0"; ), dirname $( dirname -- "$0"; )";
echo "The present working directory is $( pwd; )";
使用 pwd 单独不会工作,如果您不从它包含的目录中运行脚本。
[matt@server1 ~]$ pwd
/home/matt
[matt@server1 ~]$ ./test2.sh
The script you are running has basename test2.sh, dirname .
The present working directory is /home/matt
[matt@server1 ~]$ cd /tmp
[matt@server1 tmp]$ ~/test2.sh
The script you are running has basename test2.sh, dirname /home/matt
The present working directory is /tmp
在完整的披露中,......(我找到了这个使者的变量部分)以及在Rich’s sh的技巧中,我也在我自己的答案的披露下披露了他的页面的相关部分。
具体的:
雖然不是嚴格 POSIX 到目前為止, realpath 是一個 GNU 核心應用程式自 2012 年 全公開:我從未聽到它之前,我注意到它在 info coreutils TOC 和立即思考 [連結] 問題,但使用下列功能,如顯示應該可靠,(現在 POSIXLY?),我希望,有效地提供其呼叫者一個絕對來源的 $0:
% _abs_0() {
> o1="${1%%/*}"; ${o1:="${1}"}; ${o1:=`realpath -s "${1}"`}; eval "$1=\${o1}";
> }
% _abs_0 ${abs0:="${0}"} ; printf %s\\n "${abs0}"
/no/more/dots/in/your/path2.sh
另一方面,你可以这样做:
ps ww -fp $$ | grep -Eo '/[^:]*'"${0#*/}"
eval "abs0=${`ps ww -fp $$ | grep -Eo ' /'`#?}"
试试这:
func () {
body here
eval "$1=\${foo}"
}
foo='hello ; rm -rf /'
dest=bar
eval "$dest=$foo"
当然,下面的版本是完全安全的:
foo='hello ; rm -rf /'
dest=bar
eval "$dest=\$foo"
$0 不是一个可靠的方式来获得当前的脚本路径. 例如,这是我的.xprofile:
#!/bin/bash
echo "$0 $1 $2"
echo "${BASH_SOURCE[0]}"
# $dir/my_script.sh &
cd /tmp && ~/.xprofile && 来源 ~/.xprofile
/home/puchuu/.xprofile
/home/puchuu/.xprofile
-bash
/home/puchuu/.xprofile
因此,请使用 BASH_SURCE 代替。
我想评论上面的答案(我如何从脚本本身中获取Bash脚本的源目录?),但没有足够的声誉来做到这一点。
我在两年前在苹果的文档网站上找到了一个解决方案: https://developer.apple.com/library/archive/documentation/OpenSource/Conceptual/ShellScripting/AdvancedTechniques/AdvancedTechniques.html。
#!/bin/sh
# Get an absolute path for the poem.txt file.
POEM="$PWD/../poem.txt"
# Get an absolute path for the script file.
SCRIPT="$(which $0)"
if [ "x$(echo $SCRIPT | grep '^\/')" = "x" ] ; then
SCRIPT="$PWD/$SCRIPT"
fi
正如代码所表明的那样,一旦您获得脚本的绝对路径,则您可以使用命令命名以获得目录的路径。