我如何找到一个Bash脚本位于该脚本内部的目录的路径?

我想用Bash脚本作为另一个应用程序的启动器,我想将工作目录更改为Bash脚本所在的目录,所以我可以在该目录中的文件上运行,如下:

$ ./application

当前回答

对于具有 GNU coreutils readlink 的系统(例如,Linux):

$(readlink -f "$(dirname "$0")")

您不需要使用 BASH_SOURCE 当 $0 包含脚本文件名时。

其他回答

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

#!/usr/bin/env bash

SCRIPT_DIR=$( cd -- "$( dirname -- "${BASH_SOURCE[0]}" )" &> /dev/null && pwd )

这是一个有用的单行,这将为您提供脚本的完整目录名称,无论它从哪里被召唤。

它将工作,只要找到脚本的路径的最后一个组成部分不是一个simlink(指南链接是OK)。如果你也想解决任何链接到脚本本身,你需要一个多线解决方案:

#!/usr/bin/env bash

SOURCE=${BASH_SOURCE[0]}
while [ -L "$SOURCE" ]; do # resolve $SOURCE until the file is no longer a symlink
  DIR=$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )
  SOURCE=$(readlink "$SOURCE")
  [[ $SOURCE != /* ]] && SOURCE=$DIR/$SOURCE # if $SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
done
DIR=$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )

最后一个将与任何结合的联盟,来源,bash -c,simlinks等工作。

注意:如果您在运行此剪辑之前将CD转到另一个目录,结果可能是错误的!

此外,请注意 $CDPATH gotchas 和 stderr 输出副作用,如果用户有明智的 overridden cd 将输出转向 stderr 而不是 (包括逃避序列,如在 Mac 上呼叫 update_terminal_cwd >&2 ) 添加 >/dev/null 2>&1 在您的 cd 命令结束时,将考虑到两种可能性。

要了解它是如何工作的,试着运行这个更垂直的形式:

#!/usr/bin/env bash

SOURCE=${BASH_SOURCE[0]}
while [ -L "$SOURCE" ]; do # resolve $SOURCE until the file is no longer a symlink
  TARGET=$(readlink "$SOURCE")
  if [[ $TARGET == /* ]]; then
    echo "SOURCE '$SOURCE' is an absolute symlink to '$TARGET'"
    SOURCE=$TARGET
  else
    DIR=$( dirname "$SOURCE" )
    echo "SOURCE '$SOURCE' is a relative symlink to '$TARGET' (relative to '$DIR')"
    SOURCE=$DIR/$TARGET # if $SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
  fi
done
echo "SOURCE is '$SOURCE'"
RDIR=$( dirname "$SOURCE" )
DIR=$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )
if [ "$DIR" != "$RDIR" ]; then
  echo "DIR '$RDIR' resolves to '$DIR'"
fi
echo "DIR is '$DIR'"

它将打印一些类似:

SOURCE './scriptdir.sh' is a relative symlink to 'sym2/scriptdir.sh' (relative to '.')
SOURCE is './sym2/scriptdir.sh'
DIR './sym2' resolves to '/home/ubuntu/dotfiles/fo fo/real/real1/real2'
DIR is '/home/ubuntu/dotfiles/fo fo/real/real1/real2'

在我看来,最合适的解决方案是:

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

使用 dirname、 readlink 和 basename 最终会导致兼容性问题,所以如果可能的话最好避免。

$_ 值得提到为 $0 的替代品。 如果您正在运行来自 Bash 的脚本,可将接受的答案缩短到:

DIR="$( dirname "$_" )"

请注意,这应该是您的脚本中的第一个声明。

使用阅读链接的组合来编码名称(如果是同链接,则将其返回来源的奖金)并提取目录名称:

script="`readlink -f "${BASH_SOURCE[0]}"`"
dir="`dirname "$script"`"