没有百分之百可携带和可靠的方式来要求一个路径到当前的脚本目录,特别是在不同背景,如Cygwin,MinGW,MSYS,Linux等之间,这个问题没有正确和完全解决在Bash的年龄。

在源命令的情况下,我建议用这样的东西取代源命令:

function include()
{
  if [[ -n "$CURRENT_SCRIPT_DIR" ]]; then
    local dir_path=... get directory from `CURRENT_SCRIPT_DIR/$1`, depends if $1 is absolute path or relative ...
    local include_file_path=...
  else
    local dir_path=... request the directory from the "$1" argument using one of answered here methods...
    local include_file_path=...
  fi
  ... push $CURRENT_SCRIPT_DIR in to stack ...
  export CURRENT_SCRIPT_DIR=... export current script directory using $dir_path ...
  source "$include_file_path"
  ... pop $CURRENT_SCRIPT_DIR from stack ...
}

从现在开始,使用包括(...)是基于以前的CURRENT_SCRIPT_DIR在你的脚本。

我自己最接近这一点的实施: https://sourceforge.net/p/tacklelib/tacklelib/HEAD/tree/trunk/bash/tacklelib/bash_tacklelib https://github.com/andry81/tacklelib/tree/trunk/bash/tacklelib/bash_tacklelib

(搜索 tkl_include 函数)

总结:

FULL_PATH_TO_SCRIPT="$(realpath "${BASH_SOURCE[-1]}")"

# OR, if you do NOT need it to work for **sourced** scripts too:
# FULL_PATH_TO_SCRIPT="$(realpath "$0")"

# OR, depending on which path you want, in case of nested `source` calls
# FULL_PATH_TO_SCRIPT="$(realpath "${BASH_SOURCE[0]}")"

# OR, add `-s` to NOT expand symlinks in the path:
# FULL_PATH_TO_SCRIPT="$(realpath -s "${BASH_SOURCE[-1]}")"

SCRIPT_DIRECTORY="$(dirname "$FULL_PATH_TO_SCRIPT")"
SCRIPT_FILENAME="$(basename "$FULL_PATH_TO_SCRIPT")"

细节:

在很多情况下,所有你需要获得的是你刚刚打电话的脚本的完整路径. 这可以很容易地通过 realpath 实现. 请注意, realpath 是 GNU 核心工具的一部分. 如果你没有它已经安装(它是默认的在 Ubuntu 上),你可以安装它与 sudo apt 更新 && sudo apt 安装核心工具。

#!/bin/bash

# A. Obtain the full path, and expand (walk down) symbolic links
# A.1. `"$0"` works only if the file is **run**, but NOT if it is **sourced**.
# FULL_PATH_TO_SCRIPT="$(realpath "$0")"
# A.2. `"${BASH_SOURCE[-1]}"` works whether the file is sourced OR run, and even
# if the script is called from within another bash function!
# NB: if `"${BASH_SOURCE[-1]}"` doesn't give you quite what you want, use
# `"${BASH_SOURCE[0]}"` instead in order to get the first element from the array.
FULL_PATH_TO_SCRIPT="$(realpath "${BASH_SOURCE[-1]}")"
# B.1. `"$0"` works only if the file is **run**, but NOT if it is **sourced**.
# FULL_PATH_TO_SCRIPT_KEEP_SYMLINKS="$(realpath -s "$0")"
# B.2. `"${BASH_SOURCE[-1]}"` works whether the file is sourced OR run, and even
# if the script is called from within another bash function!
# NB: if `"${BASH_SOURCE[-1]}"` doesn't give you quite what you want, use
# `"${BASH_SOURCE[0]}"` instead in order to get the first element from the array.
FULL_PATH_TO_SCRIPT_KEEP_SYMLINKS="$(realpath -s "${BASH_SOURCE[-1]}")"

# You can then also get the full path to the directory, and the base
# filename, like this:
SCRIPT_DIRECTORY="$(dirname "$FULL_PATH_TO_SCRIPT")"
SCRIPT_FILENAME="$(basename "$FULL_PATH_TO_SCRIPT")"

# Now print it all out
echo "FULL_PATH_TO_SCRIPT = \"$FULL_PATH_TO_SCRIPT\""
echo "SCRIPT_DIRECTORY    = \"$SCRIPT_DIRECTORY\""
echo "SCRIPT_FILENAME     = \"$SCRIPT_FILENAME\""

如果您在脚本中使用“$0”而不是“${BASH_SOURCE[-1]}”,则在运行脚本时,您将获得相同的输出,而不是在提取脚本时,您将获得此不需要的输出:

~/GS/dev/eRCaGuy_hello_world/bash$ . get_script_path.sh 
FULL_PATH_TO_SCRIPT               = "/bin/bash"
SCRIPT_DIRECTORY                  = "/bin"
SCRIPT_FILENAME                   = "bash"

路径与路径之间的区别:

请注意,直路也成功地走下象征性链接来确定并指向他们的目标,而不是指向象征性链接。 如果你不想要这种行为(有时我不),然后添加到上面的直路命令,使该线看起来像这样:

# Obtain the full path, but do NOT expand (walk down) symbolic links; in
# other words: **keep** the symlinks as part of the path!
FULL_PATH_TO_SCRIPT="$(realpath -s "${BASH_SOURCE[-1]}")"

参考:

[我的答案] Unix 和 Linux:确定路径到源头 Shell 脚本

如果不是由父母脚本来源,而不是同链接,0美元就足够了:

script_path="$0"

如果源于父母脚本而不是同链接,请使用 $BASH_SOURCE 或 ${BASH_SOURCE[0]}:

script_path="$BASH_SOURCE"

如果是同链接,请使用 $BASH_SOURCE 与 realpath 或 readlink -f 获取真正的文件路径:

script_path="$(realpath "$BASH_SOURCE")"

此外,路径或 readlink -f 返回绝对路径。

要获取脚本的目录,使用 dirname:

script_directory="$(dirname "$script_path")"

笔记

对于 MacOS 而言,请在这里或在这里找到一个替代路径或阅读链接 -f. 要使代码与不为 Bash 的支柱兼容,请使用 ${var-string} 参数扩展。

使用“0”字母:

#!/usr/bin/env bash

echo "The script you are running has basename $( basename -- "$0"; ), dirname $( dirname -- "$0"; )";
echo "The present working directory is $( pwd; )";

使用 pwd 单独不会工作,如果您不从它包含的目录中运行脚本。

[matt@server1 ~]$ pwd
/home/matt
[matt@server1 ~]$ ./test2.sh
The script you are running has basename test2.sh, dirname .
The present working directory is /home/matt
[matt@server1 ~]$ cd /tmp
[matt@server1 tmp]$ ~/test2.sh
The script you are running has basename test2.sh, dirname /home/matt
The present working directory is /tmp

请尝试以下交互式解决方案:

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

如 realpath 或 readlink 等命令可能无法使用(取决于操作系统)。

注意: 在 Bash 中,建议使用 ${BASH_SOURCE[0]} 而不是 $0,否则路径在提取文件时可能会破裂(源/)。

否则,您可以在Bash中尝试下列功能:

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

这个函数需要一个论点. 如果论点已经有绝对的路径,打印它如有,否则打印 $PWD 变量 + 文件名论点(没有./ 预定)。

相关:

如何将当前的工作目录设置为Bash中的脚本目录?Bash脚本绝对路径与OS X可靠路径为Bash脚本获得完整路径

您可以使用 $BASH_SOURCE:

#!/usr/bin/env bash

scriptdir="$( dirname -- "$BASH_SOURCE"; )";

请注意,您需要使用 #!/bin/bash 而不是 #!/bin/sh 因为它是一个 Bash 扩展。