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
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
您可以做到这一点,只需将脚本名称($0)与 realpath 和/或 dirname 相结合,它适用于 Bash 和 Shell。
#!/usr/bin/env bash
RELATIVE_PATH="${0}"
RELATIVE_DIR_PATH="$(dirname "${0}")"
FULL_DIR_PATH="$(realpath "${0}" | xargs dirname)"
FULL_PATH="$(realpath "${0}")"
echo "RELATIVE_PATH->${RELATIVE_PATH}<-"
echo "RELATIVE_DIR_PATH->${RELATIVE_DIR_PATH}<-"
echo "FULL_DIR_PATH->${FULL_DIR_PATH}<-"
echo "FULL_PATH->${FULL_PATH}<-"
结果将是这样的:
# RELATIVE_PATH->./bin/startup.sh<-
# RELATIVE_DIR_PATH->./bin<-
# FULL_DIR_PATH->/opt/my_app/bin<-
# FULL_PATH->/opt/my_app/bin/startup.sh<-
$0 是脚本本身的名称
4.4 特殊变量类型
例如:LozanoMatheus/get_script_paths.sh
其他回答
总结:
FULL_PATH_TO_SCRIPT="$(realpath "${BASH_SOURCE[-1]}")"
# OR, if you do NOT need it to work for **sourced** scripts too:
# FULL_PATH_TO_SCRIPT="$(realpath "$0")"
# OR, depending on which path you want, in case of nested `source` calls
# FULL_PATH_TO_SCRIPT="$(realpath "${BASH_SOURCE[0]}")"
# OR, add `-s` to NOT expand symlinks in the path:
# FULL_PATH_TO_SCRIPT="$(realpath -s "${BASH_SOURCE[-1]}")"
SCRIPT_DIRECTORY="$(dirname "$FULL_PATH_TO_SCRIPT")"
SCRIPT_FILENAME="$(basename "$FULL_PATH_TO_SCRIPT")"
细节:
在很多情况下,所有你需要获得的是你刚刚打电话的脚本的完整路径. 这可以很容易地通过 realpath 实现. 请注意, realpath 是 GNU 核心工具的一部分. 如果你没有它已经安装(它是默认的在 Ubuntu 上),你可以安装它与 sudo apt 更新 && sudo apt 安装核心工具。
#!/bin/bash
# A. Obtain the full path, and expand (walk down) symbolic links
# A.1. `"$0"` works only if the file is **run**, but NOT if it is **sourced**.
# FULL_PATH_TO_SCRIPT="$(realpath "$0")"
# A.2. `"${BASH_SOURCE[-1]}"` works whether the file is sourced OR run, and even
# if the script is called from within another bash function!
# NB: if `"${BASH_SOURCE[-1]}"` doesn't give you quite what you want, use
# `"${BASH_SOURCE[0]}"` instead in order to get the first element from the array.
FULL_PATH_TO_SCRIPT="$(realpath "${BASH_SOURCE[-1]}")"
# B.1. `"$0"` works only if the file is **run**, but NOT if it is **sourced**.
# FULL_PATH_TO_SCRIPT_KEEP_SYMLINKS="$(realpath -s "$0")"
# B.2. `"${BASH_SOURCE[-1]}"` works whether the file is sourced OR run, and even
# if the script is called from within another bash function!
# NB: if `"${BASH_SOURCE[-1]}"` doesn't give you quite what you want, use
# `"${BASH_SOURCE[0]}"` instead in order to get the first element from the array.
FULL_PATH_TO_SCRIPT_KEEP_SYMLINKS="$(realpath -s "${BASH_SOURCE[-1]}")"
# You can then also get the full path to the directory, and the base
# filename, like this:
SCRIPT_DIRECTORY="$(dirname "$FULL_PATH_TO_SCRIPT")"
SCRIPT_FILENAME="$(basename "$FULL_PATH_TO_SCRIPT")"
# Now print it all out
echo "FULL_PATH_TO_SCRIPT = \"$FULL_PATH_TO_SCRIPT\""
echo "SCRIPT_DIRECTORY = \"$SCRIPT_DIRECTORY\""
echo "SCRIPT_FILENAME = \"$SCRIPT_FILENAME\""
如果您在脚本中使用“$0”而不是“${BASH_SOURCE[-1]}”,则在运行脚本时,您将获得相同的输出,而不是在提取脚本时,您将获得此不需要的输出:
~/GS/dev/eRCaGuy_hello_world/bash$ . get_script_path.sh
FULL_PATH_TO_SCRIPT = "/bin/bash"
SCRIPT_DIRECTORY = "/bin"
SCRIPT_FILENAME = "bash"
路径与路径之间的区别:
请注意,直路也成功地走下象征性链接来确定并指向他们的目标,而不是指向象征性链接。 如果你不想要这种行为(有时我不),然后添加到上面的直路命令,使该线看起来像这样:
# Obtain the full path, but do NOT expand (walk down) symbolic links; in
# other words: **keep** the symlinks as part of the path!
FULL_PATH_TO_SCRIPT="$(realpath -s "${BASH_SOURCE[-1]}")"
参考:
[我的答案] Unix 和 Linux:确定路径到源头 Shell 脚本
最短、最优雅的方式就是:
#!/bin/bash
DIRECTORY=$(cd `dirname $0` && pwd)
echo $DIRECTORY
它会在所有平台上工作,而且非常干净。
更多详细信息可在“该 bash 脚本在哪个目录?”中找到。
$_ 值得提到为 $0 的替代品。 如果您正在运行来自 Bash 的脚本,可将接受的答案缩短到:
DIR="$( dirname "$_" )"
请注意,这应该是您的脚本中的第一个声明。
在我看来,最合适的解决方案是:
"$( cd "$( echo "${BASH_SOURCE[0]%/*}" )"; pwd )"
使用 dirname、 readlink 和 basename 最终会导致兼容性问题,所以如果可能的话最好避免。
命名命令是最基本的,简单地将路径到0美元(脚本名称)变量的名称:
dirname -- "$0";
但是,正如Matt b指出的那样,返回的路径取决于脚本的名称。 pwd 不做工作,因为它只告诉你当前的目录是什么,而不是脚本的目录是什么。
有些人提到了阅读链接命令,但最简单的是,你可以使用:
dirname -- "$( readlink -f -- "$0"; )";
阅读链接将解决脚本路径从文件系统的根源到绝对路径,因此,任何包含单点或双点的路径,图标和/或象征性链接将解决到完整路径。
#!/usr/bin/env bash
echo "pwd: `pwd`"
echo "\$0: $0"
echo "basename: `basename -- "$0"`"
echo "dirname: `dirname -- "$0"`"
echo "dirname/readlink: $( dirname -- "$( readlink -f -- "$0"; )"; )"
在我的家中运行这个脚本,使用相对的路径:
>>>$ ./whatdir.sh
pwd: /Users/phatblat
$0: ./whatdir.sh
basename: whatdir.sh
dirname: .
dirname/readlink: /Users/phatblat
再一次,但使用完整的路径到脚本:
>>>$ /Users/phatblat/whatdir.sh
pwd: /Users/phatblat
$0: /Users/phatblat/whatdir.sh
basename: whatdir.sh
dirname: /Users/phatblat
dirname/readlink: /Users/phatblat
现在更改目录:
>>>$ cd /tmp
>>>$ ~/whatdir.sh
pwd: /tmp
$0: /Users/phatblat/whatdir.sh
basename: whatdir.sh
dirname: /Users/phatblat
dirname/readlink: /Users/phatblat
最后,使用一个象征性的链接来执行脚本:
>>>$ ln -s ~/whatdir.sh whatdirlink.sh
>>>$ ./whatdirlink.sh
pwd: /tmp
$0: ./whatdirlink.sh
basename: whatdirlink.sh
dirname: .
dirname/readlink: /Users/phatblat
然而,有一個案例,這不起作用,當脚本來源(而不是執行)在 bash:
>>>$ cd /tmp
>>>$ . ~/whatdir.sh
pwd: /tmp
$0: bash
basename: bash
dirname: .
dirname/readlink: /tmp