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

简单:

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

不需要花哨的运算符。

2013-12-03 13:59:03

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?

2013-10-04 11:56:30

只是为了它的地狱，我做了一些黑客在一个脚本上做的事情，纯粹的文本，纯粹的Bash。我希望我掌握了所有的边缘情况。

注意，我在另一个答案中提到的${var//pat/repl}不起作用，因为你不能让它只替换最短的匹配，这是替换/foo/的一个问题。/例如/*/../将接受它前面的所有内容，而不仅仅是一个条目。由于这些模式并不是真正的正则表达式，我不知道如何才能使其工作。这就是我想出的巧妙的解决方案，请欣赏。；）

顺便说一句，如果你发现任何未处理的边缘情况，请告诉我。

#!/bin/bash

canonicalize_path() {
  local path="$1"
  OIFS="$IFS"
  IFS=$'/'
  read -a parts < <(echo "$path")
  IFS="$OIFS"

  local i=${#parts[@]}
  local j=0
  local back=0
  local -a rev_canon
  while (($i > 0)); do
    ((i--))
    case "${parts[$i]}" in
      ""|.) ;;
      ..) ((back++));;
      *) if (($back > 0)); then
           ((back--))
         else
           rev_canon[j]="${parts[$i]}"
           ((j++))
         fi;;
    esac
  done
  while (($j > 0)); do
    ((j--))
    echo -n "/${rev_canon[$j]}"
  done
  echo
}

canonicalize_path "/.././..////../foo/./bar//foo/bar/.././bar/../foo/bar/./../..//../foo///bar/"

2011-01-23 14:53:44

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)

2014-04-18 20:05:31

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)。

Bash脚本获取自身完整路径的可靠方法

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