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

您可以尝试定义以下变量:

CWD="$(cd -P -- "$(dirname -- "${BASH_SOURCE[0]}")" && pwd -P)"

或者你可以在Bash中尝试以下函数:

realpath () {
  [[ $1 = /* ]] && echo "$1" || echo "$PWD/${1#./}"
}

这个函数有一个参数。如果参数已经有一个绝对路径,则打印它,否则打印$PWD变量+文件名参数(不带。/前缀)。

相关:

Bash脚本绝对路径与OS X 从脚本本身中获取Bash脚本的源目录

其他回答

我很惊讶这里没有提到realpath命令。我的理解是它可以广泛移植。

你的初始解决方案是:

SCRIPT=$(realpath "$0")
SCRIPTPATH=$(dirname "$SCRIPT")

并根据您的喜好留下未解决的符号链接:

SCRIPT=$(realpath -s "$0")
SCRIPTPATH=$(dirname "$SCRIPT")

Use:

SCRIPT_PATH=$(dirname `which $0`)

它将可执行文件的完整路径打印到标准输出,该路径是在shell提示符下输入传入参数时执行的($0包含该参数)

Dirname从文件名中去掉非目录后缀。

因此,无论是否指定了路径,您最终都会得到脚本的完整路径。

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)

我今天不得不重新讨论这个问题,并从脚本本身中找到了获取Bash脚本的源目录:

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

在链接的答案中有更多的变体,例如在脚本本身是符号链接的情况下。

还有另一种方法:

shopt -s extglob

selfpath=$0
selfdir=${selfpath%%+([!/])}

while [[ -L "$selfpath" ]];do
  selfpath=$(readlink "$selfpath")
  if [[ ! "$selfpath" =~ ^/ ]];then
    selfpath=${selfdir}${selfpath}
  fi
  selfdir=${selfpath%%+([!/])}
done

echo $selfpath $selfdir