Bash脚本获取自身完整路径的可靠方法

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).

当前回答

我发现在Bash中获得完整规范路径的最简单方法是使用cd和pwd:

ABSOLUTE_PATH="$(cd "$(dirname "${BASH_SOURCE[0]}")" && pwd)/$(basename "${BASH_SOURCE[0]}")"

使用${BASH_SOURCE[0]}而不是$0，无论脚本是作为<name>还是source <name>调用，都会产生相同的行为。

2012-02-02 04:03:51

其他回答

简单:

BASEDIR=$(readlink -f $0 | xargs dirname)

不需要花哨的运算符。

2013-12-03 13:59:03

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。

2011-01-23 13:25:57

更简单地说，这对我来说是有效的:

MY_DIR=`dirname $0`
source $MY_DIR/_inc_db.sh

2016-11-15 11:31:48

易于阅读?下面是一个替代方案。它忽略了符号链接

#!/bin/bash
currentDir=$(
  cd $(dirname "$0")
  pwd
)

echo -n "current "
pwd
echo script $currentDir

自从几年前我发布了上面的答案，我已经发展了我的实践，使用这个linux特定的范例，它正确地处理符号链接:

ORIGIN=$(dirname $(readlink -f $0))

2016-05-12 18:43:39

我已经成功地使用了下面的方法一段时间(不是在OS X上)，它只使用一个内置的shell，并处理'source foobar.sh'的情况，就我所见。

下面的示例代码的一个问题是，函数使用$PWD，在函数调用时，$PWD可能正确，也可能不正确。所以这需要处理。

#!/bin/bash

function canonical_path() {
  # Handle relative vs absolute path
  [ ${1:0:1} == '/' ] && x=$1 || x=$PWD/$1
  # Change to dirname of x
  cd ${x%/*}
  # Combine new pwd with basename of x
  echo $(pwd -P)/${x##*/}
  cd $OLDPWD
}

echo $(canonical_path "${BASH_SOURCE[0]}")

type [
type cd
type echo
type pwd

2014-04-18 21:38:57

Bash脚本获取自身完整路径的可靠方法

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