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

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)"

BASE_DIR="$(cd "$(dirname "$0")"; pwd)";
echo "BASE_DIR => $BASE_DIR"

正如marko所言:

BASEDIR=$(dirname $0)
echo $BASEDIR

除非你在脚本所在的目录中执行脚本，在这种情况下，你会得到一个值'。'

要解决这个问题，请使用:

current_dir=$(pwd)
script_dir=$(dirname $0)

if [ $script_dir = '.' ]
then
script_dir="$current_dir"
fi

现在可以在整个脚本中使用变量current_dir来引用脚本目录。然而，这可能仍然有符号链接的问题。

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