这个问题的最佳答案是: 从内部获取Bash脚本的源目录

它是:

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

一行代码，它将提供脚本的完整目录名，无论从哪里调用脚本。

要了解它是如何工作的，你可以执行以下脚本:

#!/bin/bash

SOURCE="${BASH_SOURCE[0]}"
while [ -h "$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 "$SOURCE" )" && pwd )"
if [ "$DIR" != "$RDIR" ]; then
  echo "DIR '$RDIR' resolves to '$DIR'"
fi
echo "DIR is '$DIR'"

这应该能奏效:

echo `pwd`/`dirname $0`

它可能看起来很丑，这取决于它是如何被调用和cwd，但应该得到你需要去的地方(或者你可以调整字符串，如果你关心它的外观)。

灵感来自blueyed的回答

read < <(readlink -f $0 | xargs dirname)
cd $REPLY

如果你想获得实际的脚本目录(不管你是使用符号链接还是直接调用脚本)，尝试:

BASEDIR=$(dirname $(realpath "$0"))
echo "$BASEDIR"

这在linux和macOS上都适用。我没看到有人提到realpath。不确定这种方法是否有任何缺点。

在macOS上，需要安装coreutils才能使用realpath。例如:brew install coreutils。

So many answers, all plausible, each with pro's and con's & slightly differeing objectives (which should probably be stated for each). Here's another solution that meets a primary objective of both being clear and working across all systems, on all bash (no assumptions about bash versions, or readlink or pwd options), and reasonably does what you'd expect to happen (eg, resolving symlinks is an interesting problem, but isn't usually what you actually want), handle edge cases like spaces in paths, etc., ignores any errors and uses a sane default if there are any issues.

每个组件都存储在一个单独的变量中，您可以单独使用:

# script path, filename, directory
PROG_PATH=${BASH_SOURCE[0]}      # this script's name
PROG_NAME=${PROG_PATH##*/}       # basename of script (strip path)
PROG_DIR="$(cd "$(dirname "${PROG_PATH:-$PWD}")" 2>/dev/null 1>&2 && pwd)"

最初的帖子包含了解决方案(忽略回复，他们没有添加任何有用的东西)。有趣的工作是由前面提到的unix命令readlink和选项-f完成的。当以绝对路径和相对路径调用脚本时，此方法有效。

对于bash, sh, ksh:

#!/bin/bash 
# Absolute path to this script, e.g. /home/user/bin/foo.sh
SCRIPT=$(readlink -f "$0")
# Absolute path this script is in, thus /home/user/bin
SCRIPTPATH=$(dirname "$SCRIPT")
echo $SCRIPTPATH

对于tcsh, csh:

#!/bin/tcsh
# Absolute path to this script, e.g. /home/user/bin/foo.csh
set SCRIPT=`readlink -f "$0"`
# Absolute path this script is in, thus /home/user/bin
set SCRIPTPATH=`dirname "$SCRIPT"`
echo $SCRIPTPATH

参见:https://stackoverflow.com/a/246128/59087