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

没有<unk>(除了<unk>)和可以处理“陌生人”名称的形式,如那些有新闻,因为有些人会声称:

IFS= read -rd '' DIR < <([[ $BASH_SOURCE != */* ]] || cd "${BASH_SOURCE%/*}/" >&- && echo -n "$PWD")

其他回答

这在Bash 3.2中工作:

path="$( dirname "$( which "$0" )" )"

如果您在 $PATH 中有 ~/bin 目录,则在此目录中有 A. 它源于 ~/bin/lib/B. 您知道所包含的目录与原始目录相对,在 lib 子目录中,但不是与用户当前目录相对。

以以下方式解决问题(A内部):

source "$( dirname "$( which "$0" )" )/lib/B"

无论用户在哪里,还是他/她如何呼叫脚本,这总是会工作。

下面将返回剧本的当前目录

工作,如果它是源,或者不源工作,如果运行在当前的目录,或某些其他目录.工作,如果相对目录被使用.工作与 bash,不确定其他<unk>。

/tmp/a/b/c $ . ./test.sh
/tmp/a/b/c

/tmp/a/b/c $ . /tmp/a/b/c/test.sh
/tmp/a/b/c

/tmp/a/b/c $ ./test.sh
/tmp/a/b/c

/tmp/a/b/c $ /tmp/a/b/c/test.sh
/tmp/a/b/c

/tmp/a/b/c $ cd

~ $ . /tmp/a/b/c/test.sh
/tmp/a/b/c

~ $ . ../../tmp/a/b/c/test.sh
/tmp/a/b/c

~ $ /tmp/a/b/c/test.sh
/tmp/a/b/c

~ $ ../../tmp/a/b/c/test.sh
/tmp/a/b/c

测试.sh

#!/usr/bin/env bash

# snagged from: https://stackoverflow.com/a/51264222/26510
function toAbsPath {
    local target
    target="$1"

    if [ "$target" == "." ]; then
        echo "$(pwd)"
    elif [ "$target" == ".." ]; then
        echo "$(dirname "$(pwd)")"
    else
        echo "$(cd "$(dirname "$1")"; pwd)/$(basename "$1")"
    fi
}

function getScriptDir(){
  local SOURCED
  local RESULT
  (return 0 2>/dev/null) && SOURCED=1 || SOURCED=0

  if [ "$SOURCED" == "1" ]
  then
    RESULT=$(dirname "$1")
  else
    RESULT="$( cd "$( dirname "${BASH_SOURCE[0]}" )" >/dev/null 2>&1 && pwd )"
  fi
  toAbsPath "$RESULT"
}

SCRIPT_DIR=$(getScriptDir "$0")
echo "$SCRIPT_DIR"
#!/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'

我会用这样的东西:

# Retrieve the full pathname of the called script
scriptPath=$(which $0)

# Check whether the path is a link or not
if [ -L $scriptPath ]; then

    # It is a link then retrieve the target path and get the directory name
    sourceDir=$(dirname $(readlink -f $scriptPath))

else

    # Otherwise just get the directory name of the script path
    sourceDir=$(dirname $scriptPath)

fi

关键部分是,我正在减少问题的范围:我禁止通过路径间接执行脚本(如 /bin/sh [脚本路径与路径组件有关])。

这可以被检测到,因为0美元将是一个相对的路径,不解决与当前文件夹有关的任何文件。我相信使用#!机制的直接执行总是导致绝对0美元,包括当脚本在路径上找到时。

我也要求在象征性链接链接链接的任何字符和字符只包含一个合理的字符子,特别是不是 \n, >, * 或?. 这对于字符逻辑来说是必要的。

#!/bin/sh
(
    path="${0}"
    while test -n "${path}"; do
        # Make sure we have at least one slash and no leading dash.
        expr "${path}" : / > /dev/null || path="./${path}"
        # Filter out bad characters in the path name.
        expr "${path}" : ".*[*?<>\\]" > /dev/null && exit 1
        # Catch embedded new-lines and non-existing (or path-relative) files.
        # $0 should always be absolute when scripts are invoked through "#!".
        test "`ls -l -d "${path}" 2> /dev/null | wc -l`" -eq 1 || exit 1
        # Change to the folder containing the file to resolve relative links.
        folder=`expr "${path}" : "\(.*/\)[^/][^/]*/*$"` || exit 1
        path=`expr "x\`ls -l -d "${path}"\`" : "[^>]* -> \(.*\)"`
        cd "${folder}"
        # If the last path was not a link then we are in the target folder.
        test -n "${path}" || pwd
    done
)