我有一个Bash脚本，需要知道它的完整路径。我试图找到一种广泛兼容的方式来做到这一点，而不会以相对或时髦的路径结束。我只需要支持Bash，不支持sh, csh等。
到目前为止，我发现:
The accepted answer to Getting the source directory of a Bash script from within addresses getting the path of the script via dirname $0, which is fine, but that may return a relative path (like .), which is a problem if you want to change directories in the script and have the path still point to the script's directory. Still, dirname will be part of the puzzle.
The accepted answer to Bash script absolute path with OS X (OS X specific, but the answer works regardless) gives a function that will test to see if $0 looks relative and if so will pre-pend $PWD to it. But the result can still have relative bits in it (although overall it's absolute) — for instance, if the script is t in the directory /usr/bin and you're in /usr and you type bin/../bin/t to run it (yes, that's convoluted), you end up with /usr/bin/../bin as the script's directory path. Which works, but...
The readlink solution on this page, which looks like this:
# Absolute path to this script. /home/user/bin/foo.sh
SCRIPT=$(readlink -f $0)
# Absolute path this script is in. /home/user/bin
SCRIPTPATH=`dirname $SCRIPT`
But readlink isn't POSIX and apparently the solution relies on GNU's readlink where BSD's won't work for some reason (I don't have access to a BSD-like system to check).
有很多种方法，但都有注意事项。
还有什么更好的办法呢?“更好”的意思是:
Gives me the absolute path.
Takes out funky bits even when invoked in a convoluted way (see comment on #2 above). (E.g., at least moderately canonicalizes the path.)
Relies only on Bash-isms or things that are almost certain to be on most popular flavors of *nix systems (GNU/Linux, BSD and BSD-like systems like OS X, etc.).
Avoids calling external programs if possible (e.g., prefers Bash built-ins).
(Updated, thanks for the heads up, wich) It doesn't have to resolve symlinks (in fact, I'd kind of prefer it left them alone, but that's not a requirement).
以下是我所想到的(编辑:加上一些由sfstewman, levigroker, Kyle Strand和Rob Kennedy提供的调整)，似乎基本上符合我的“更好”标准:
SCRIPTPATH="$( cd -- "$(dirname "$0")" >/dev/null 2>&1 ; pwd -P )"
SCRIPTPATH行似乎特别迂回，但为了正确地处理空格和符号链接，我们需要它而不是SCRIPTPATH= ' pwd '。
包含输出重定向(>/dev/null 2>&1)可以处理罕见的(?)情况，即cd可能产生的输出会干扰周围的$(…)捕捉。(例如cd被覆盖，也ls一个目录后切换到它。)
还要注意一些深奥的情况，比如执行一个根本不是来自可访问文件系统中的文件的脚本(这是完全可能的)，不适合在那里(或者在我看到的任何其他答案中)。
cd后面和“$0”之前的——是为了防止目录以-开头。
被接受的解决方案(对我来说)不方便“来源”:
如果你从“来源../..”/yourScript"， $0将是"bash"!
下面的函数(对于bash >= 3.0)给出了正确的路径，但是脚本可能会被调用(直接或通过源代码，使用绝对路径或相对路径):
(这里的“正确路径”指的是被调用脚本的完整绝对路径，即使是从另一个路径直接调用，也可以使用“source”)
#!/bin/bash
echo $0 executed
function bashscriptpath() {
local _sp=$1
local ascript="$0"
local asp="$(dirname $0)"
#echo "b1 asp '$asp', b1 ascript '$ascript'"
if [[ "$asp" == "." && "$ascript" != "bash" && "$ascript" != "./.bashrc" ]] ; then asp="${BASH_SOURCE[0]%/*}"
elif [[ "$asp" == "." && "$ascript" == "./.bashrc" ]] ; then asp=$(pwd)
else
if [[ "$ascript" == "bash" ]] ; then
ascript=${BASH_SOURCE[0]}
asp="$(dirname $ascript)"
fi
#echo "b2 asp '$asp', b2 ascript '$ascript'"
if [[ "${ascript#/}" != "$ascript" ]]; then asp=$asp ;
elif [[ "${ascript#../}" != "$ascript" ]]; then
asp=$(pwd)
while [[ "${ascript#../}" != "$ascript" ]]; do
asp=${asp%/*}
ascript=${ascript#../}
done
elif [[ "${ascript#*/}" != "$ascript" ]]; then
if [[ "$asp" == "." ]] ; then asp=$(pwd) ; else asp="$(pwd)/${asp}"; fi
fi
fi
eval $_sp="'$asp'"
}
bashscriptpath H
export H=${H}
关键是检测“source”大小写，并使用${BASH_SOURCE[0]}返回实际的脚本。
获取shell脚本的绝对路径
它没有在readlink中使用-f选项，因此它应该在BSD/Mac OS X上工作。
支持
source ./script(当被。点运算符)
绝对路径/path/to/script
相对路径，比如。/script
/道路/ dir1 / . . / dir2 / dir3 / . . /脚本
当从symlink调用时
当符号链接嵌套eg) foo->dir1/dir2/bar bar->./../能源部doe - >脚本
当调用者更改脚本名称时
我正在寻找这段代码不能工作的极端情况。请让我知道。
Code
pushd . > /dev/null
SCRIPT_PATH="${BASH_SOURCE[0]}";
while([ -h "${SCRIPT_PATH}" ]); do
cd "`dirname "${SCRIPT_PATH}"`"
SCRIPT_PATH="$(readlink "`basename "${SCRIPT_PATH}"`")";
done
cd "`dirname "${SCRIPT_PATH}"`" > /dev/null
SCRIPT_PATH="`pwd`";
popd > /dev/null
echo "srcipt=[${SCRIPT_PATH}]"
echo "pwd =[`pwd`]"
已知的政务
脚本必须在磁盘的某个地方。让它通过网络。如果您试图从PIPE运行这个脚本，它将无法工作
wget -o /dev/null -O - http://host.domain/dir/script.sh |bash
从技术上讲，它是没有定义的。实际上，没有明智的方法来检测这一点。(协进程不能访问父进程的环境。)
我们在GitHub上放置了自己的产品realpath-lib，供社区免费使用。
无耻的插头，但有了这个Bash库，你可以:
get_realpath <absolute|relative|symlink|local file>
这个函数是库的核心:
function get_realpath() {
if [[ -f "$1" ]]
then
# file *must* exist
if cd "$(echo "${1%/*}")" &>/dev/null
then
# file *may* not be local
# exception is ./file.ext
# try 'cd .; cd -;' *works!*
local tmppwd="$PWD"
cd - &>/dev/null
else
# file *must* be local
local tmppwd="$PWD"
fi
else
# file *cannot* exist
return 1 # failure
fi
# reassemble realpath
echo "$tmppwd"/"${1##*/}"
return 0 # success
}
它不需要任何外部依赖，只需要Bash 4+。还包含函数get_dirname, get_filename, get_stemname和validate_path validate_realpath。它是免费的，干净的，简单的，有良好的文档，所以它也可以用于学习目的，毫无疑问，它是可以改进的。尝试跨平台。
更新:经过一些审查和测试，我们已经将上面的函数替换为可以达到相同结果的函数(没有使用dirname，只使用纯Bash)，但效率更高:
function get_realpath() {
[[ ! -f "$1" ]] && return 1 # failure : file does not exist.
[[ -n "$no_symlinks" ]] && local pwdp='pwd -P' || local pwdp='pwd' # do symlinks.
echo "$( cd "$( echo "${1%/*}" )" 2>/dev/null; $pwdp )"/"${1##*/}" # echo result.
return 0 # success
}
这还包括一个环境设置no_symlinks，它提供了将符号链接解析到物理系统的能力。默认情况下，它保持符号链接不变。
