这一行代码告诉shell脚本在哪里，与您是否运行它或是否获取它无关。此外，它还会解析所涉及的任何符号链接，如果是这样的话:

dir=$(dirname $(test -L "$BASH_SOURCE" && readlink -f "$BASH_SOURCE" || echo "$BASH_SOURCE"))

顺便说一下，我认为您正在使用/bin/bash.

灵感来自blueyed的回答

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

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)。

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'"

之前对一个答案的评论说过，但在所有其他答案中很容易被忽略。

使用bash时:

echo this file: "$BASH_SOURCE"
echo this dir: "$(dirname "$BASH_SOURCE")"

Bash参考手册，5.2 Bash变量

基本版:

dir=$(dirname $0)

如果脚本可以通过$PATH调用，那么:

dir=$(dirname $(which $0))

如果脚本可能像这样调用:bash script.sh，那么:

dir=$(dirname $(which $0 2>/dev/null || realpath ./$0))

如果你感到极度不安全，那么:

dir="$(dirname -- "$(which -- "$0" 2>/dev/null || realpath -- "./$0")")"