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

由于realpath没有按默认安装在我的Linux系统上,下面的工作为我:

SCRIPT="$(readlink --canonicalize-existing "$0")"
SCRIPTPATH="$(dirname "$SCRIPT")"

$SCRIPT将包含脚本的真实文件路径,$SCRIPTPATH将包含脚本的目录的真实路径。

在使用这个答案之前,请阅读这个答案的注释。

其他回答

再次考虑这个问题:在这个线程中引用了一个非常流行的解决方案,它的起源在这里:

DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )"

我一直没有使用这种解决方案,因为使用了dirname——它可能会带来跨平台的困难,特别是在出于安全原因需要锁定脚本的情况下。但是作为一个纯Bash的替代品,如何使用:

DIR="$( cd "$( echo "${BASH_SOURCE[0]%/*}" )" && pwd )"

这是一个选择吗?

如果我们使用Bash,我相信这是最方便的方式,因为它不需要调用任何外部命令:

THIS_PATH="${BASH_SOURCE[0]}";
THIS_DIR=$(dirname $THIS_PATH)

Bourne shell (sh)兼容方式:

SCRIPT_HOME=`dirname $0 | while read a; do cd $a && pwd && break; done`

回答这个问题很晚,但我用:

SCRIPT=$( readlink -m $( type -p ${0} ))      # Full path to script handling Symlinks
BASE_DIR=`dirname "${SCRIPT}"`                # Directory script is run in
NAME=`basename "${SCRIPT}"`                   # Actual name of script even if linked

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))