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

使用“0”字母:

#!/usr/bin/env bash

echo "The script you are running has basename $( basename -- "$0"; ), dirname $( dirname -- "$0"; )";
echo "The present working directory is $( pwd; )";

使用 pwd 单独不会工作,如果您不从它包含的目录中运行脚本。

[matt@server1 ~]$ pwd
/home/matt
[matt@server1 ~]$ ./test2.sh
The script you are running has basename test2.sh, dirname .
The present working directory is /home/matt
[matt@server1 ~]$ cd /tmp
[matt@server1 tmp]$ ~/test2.sh
The script you are running has basename test2.sh, dirname /home/matt
The present working directory is /tmp

其他回答

我想确保剧本在其目录中运行。

cd $(dirname $(which $0) )

接下来,如果你真的想知道你在哪里运行,然后运行下面的命令。

DIR=$(/usr/bin/pwd)

我认为最简单的答案是原始变量的参数扩展:

#!/usr/bin/env bash

DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" >/dev/null 2>&1 && pwd )"
echo "opt1; original answer: $DIR"
echo ''

echo "opt2; simple answer  : ${BASH_SOURCE[0]%/*}"

它应该产生产量如:

$ /var/tmp/test.sh
opt1; original answer: /var/tmp

opt2; simple answer  : /var/tmp

变量/参数扩展 ${BASH_SOURCE[0]%/*}”似乎更容易保持。

我尝试了所有这些,没有一个工作,一个非常接近,但有一个小错误打破了它;他们忘了把路放在引用标志上。

而且,很多人假设你正在从一个阴影中运行脚本,所以当你打开一个新的脚本时,他们会忘记它到你的家。

请尝试这个目录为尺寸:

/var/No one/Thought/About Spaces Being/In a Directory/Name/And Here's your file.text

这是正确的,无论你如何或在哪里运行它:

#!/bin/bash
echo "pwd: `pwd`"
echo "\$0: $0"
echo "basename: `basename "$0"`"
echo "dirname: `dirname "$0"`"

因此,要使它实际上有用,这里是如何改变运行脚本的目录:

cd "`dirname "$0"`"

使用“0”字母:

#!/usr/bin/env bash

echo "The script you are running has basename $( basename -- "$0"; ), dirname $( dirname -- "$0"; )";
echo "The present working directory is $( pwd; )";

使用 pwd 单独不会工作,如果您不从它包含的目录中运行脚本。

[matt@server1 ~]$ pwd
/home/matt
[matt@server1 ~]$ ./test2.sh
The script you are running has basename test2.sh, dirname .
The present working directory is /home/matt
[matt@server1 ~]$ cd /tmp
[matt@server1 tmp]$ ~/test2.sh
The script you are running has basename test2.sh, dirname /home/matt
The present working directory is /tmp
#!/bin/sh
PRG="$0"

# need this for relative symlinks
while [ -h "$PRG" ] ; do
   PRG=`readlink "$PRG"`
done

scriptdir=`dirname "$PRG"`