我有一个脚本,我不希望它调用退出,如果它是来源。
我想检查是否$0 == bash,但这有问题,如果脚本是从另一个脚本,或者如果用户从不同的shell,如ksh源。
是否有一种可靠的方法来检测脚本是否被引用?
我有一个脚本,我不希望它调用退出,如果它是来源。
我想检查是否$0 == bash,但这有问题,如果脚本是从另一个脚本,或者如果用户从不同的shell,如ksh源。
是否有一种可靠的方法来检测脚本是否被引用?
当前回答
检测Bash脚本是否正在执行或导入的最漂亮方法
我真的认为这是最美丽的方式:
从我的eRCaGuy_hello_world repo中的if__name__==__main__ check_if_sourced_or_executed_best.sh文件:
#!/usr/bin/env bash
main() {
echo "Running main."
# Add your main function code here
}
if [ "${BASH_SOURCE[0]}" = "$0" ]; then
# This script is being run.
__name__="__main__"
else
# This script is being sourced.
__name__="__source__"
fi
# Only run `main` if this script is being **run**, NOT sourced (imported)
if [ "$__name__" = "__main__" ]; then
echo "This script is being run."
main
else
echo "This script is being sourced."
fi
引用:
关于上述技术的其他详细信息,请参见我在这里的另一个回答,包括显示运行输出:bash与Python的if __name__ == '__main__'等价于什么? 这个答案,我第一次了解到“${BASH_SOURCE[0]}”=“$0”
如果您愿意,还可以探索以下替代方案,但我更喜欢使用上面的代码块。
重要提示:使用“${FUNCNAME[-1]}”技术不能正确处理嵌套脚本,即一个脚本调用或来源另一个脚本,而if ["${BASH_SOURCE[0]}" = "$0"]技术可以。这是使用if ["${BASH_SOURCE[0]}" = "$0"]的另一个重要原因。
4种方法确定bash脚本是源脚本还是执行脚本
我已经阅读了关于这个问题和其他一些问题的一堆答案,并提出了4种我想要总结并放在一个地方的方法。
if __name__ == "__main__":
参见:如果__name__ == "__main__":会做什么?在Python中所做的事情。
You can see a full demonstration of all 4 techniques below in my check_if_sourced_or_executed.sh script in my eRCaGuy_hello_world repo. You can see one of the techniques in-use in my advanced bash program with help menu, argument parsing, main function, automatic execute vs source detection (akin to if __name__ == "__main__": in Python), etc, see my demo/template program in this list here. It is currently called argument_parsing__3_advanced__gen_prog_template.sh, but if that name changes in the future I'll update it in the list at the link just above
不管怎样,这里有4个Bash技术:
Technique 1 (can be placed anywhere; handles nested scripts): See: https://unix.stackexchange.com/questions/424492/how-to-define-a-shell-script-to-be-sourced-not-run/424495#424495 if [ "${BASH_SOURCE[0]}" -ef "$0" ]; then echo " This script is being EXECUTED." run="true" else echo " This script is being SOURCED." fi Technique 2 [My favorite technique] (can be placed anywhere; handles nestes scripts): See this type of technique in-use in my most-advanced bash demo script yet, here: argument_parsing__3_advanced__gen_prog_template.sh, near the bottom. Modified from: What is the bash equivalent to Python's `if __name__ == '__main__'`? if [ "${BASH_SOURCE[0]}" == "$0" ]; then echo " This script is being EXECUTED." run="true" else echo " This script is being SOURCED." fi Technique 3 (requires another line which MUST be outside all functions): Modified from: How to detect if a script is being sourced # A. Place this line OUTSIDE all functions: (return 0 2>/dev/null) && script_is_being_executed="false" || script_is_being_executed="true" # B. Place these lines anywhere if [ "$script_is_being_executed" == "true" ]; then echo " This script is being EXECUTED." run="true" else echo " This script is being SOURCED." fi Technique 4 [Limitation: does not handle nested scripts!] (MUST be inside a function): Modified from: How to detect if a script is being sourced and Unix & Linux: How to define a shell script to be sourced not run. if [ "${FUNCNAME[-1]}" == "main" ]; then echo " This script is being EXECUTED." run="true" elif [ "${FUNCNAME[-1]}" == "source" ]; then echo " This script is being SOURCED." else echo " ERROR: THIS TECHNIQUE IS BROKEN" fi This is where I first learned about the ${FUNCNAME[-1]} trick: @mr.spuratic: How to detect if a script is being sourced - he learned it from Dennis Williamson apparently.
参见:
[我的回答]bash相当于Python的if __name__ == '__main__'? [我的回答]Unix和Linux:如何定义一个shell脚本来获取而不是运行
其他回答
这是从其他一些关于“通用”跨壳支持的答案衍生出来的。不可否认,这与https://stackoverflow.com/a/2942183/3220983非常相似,尽管略有不同。这样做的缺点是,客户端脚本必须尊重如何使用它(即先导出一个变量)。它的优点是简单,而且可以在“任何地方”工作。这里有一个模板供你剪切和粘贴:
# NOTE: This script may be used as a standalone executable, or callable library.
# To source this script, add the following *prior* to including it:
# export ENTRY_POINT="$0"
main()
{
echo "Running in direct executable context!"
}
if [ -z "${ENTRY_POINT}" ]; then main "$@"; fi
注意:我使用export只是为了确保这个机制可以扩展到子进程。
如果您的Bash版本知道BASH_SOURCE数组变量,请尝试如下操作:
# man bash | less -p BASH_SOURCE
#[[ ${BASH_VERSINFO[0]} -le 2 ]] && echo 'No BASH_SOURCE array variable' && exit 1
[[ "${BASH_SOURCE[0]}" != "${0}" ]] && echo "script ${BASH_SOURCE[0]} is being sourced ..."
解决这个问题的方法不是编写需要知道这些事情才能正确运行的代码。做到这一点的方法是将代码放入函数中,而不是放入需要源代码的脚本主线中。
函数内部的代码可以只返回0或1。这只终止了函数,因此控制返回到调用该函数的任何对象。
无论从源脚本的主线调用函数,还是从顶级脚本的主线调用函数,或者从另一个函数调用函数,都是如此。
使用sourcing来引入“库”脚本,这些脚本只定义函数和变量,但实际上不执行任何其他顶级命令:
. path/to/lib.sh # defines libfunction
libfunction arg
否则:
path/to/script.sh arg # call script as a child process
而不是:
. path/to/script.sh arg # shell programming anti-pattern
$_是很脆弱的。你必须检查它作为你在脚本中做的第一件事。即使这样,它也不保证包含shell的名称(如果是源的)或脚本的名称(如果是执行的)。
例如,如果用户设置了BASH_ENV,那么在脚本的顶部,$_包含BASH_ENV脚本中执行的最后一个命令的名称。
我发现最好的方法是像这样使用0美元:
name="myscript.sh"
main()
{
echo "Script was executed, running main..."
}
case "$0" in *$name)
main "$@"
;;
esac
不幸的是,这种方式在zsh中并不能开箱使用,因为functionargzero选项的功能超出了它的名称,并且在默认情况下是打开的。
为了解决这个问题,我把unsetopt functionarg0放在我的.zshenv中。
我将给出一个特定于bash的答案。Korn shell,对不起。假设脚本名为include2.sh;然后在include2.sh中创建一个名为am_I_sourced的函数。下面是我的include2.sh的演示版本:
am_I_sourced()
{
if [ "${FUNCNAME[1]}" = source ]; then
if [ "$1" = -v ]; then
echo "I am being sourced, this filename is ${BASH_SOURCE[0]} and my caller script/shell name was $0"
fi
return 0
else
if [ "$1" = -v ]; then
echo "I am not being sourced, my script/shell name was $0"
fi
return 1
fi
}
if am_I_sourced -v; then
echo "Do something with sourced script"
else
echo "Do something with executed script"
fi
现在尝试以多种方式执行它:
~/toys/bash $ chmod a+x include2.sh
~/toys/bash $ ./include2.sh
I am not being sourced, my script/shell name was ./include2.sh
Do something with executed script
~/toys/bash $ bash ./include2.sh
I am not being sourced, my script/shell name was ./include2.sh
Do something with executed script
~/toys/bash $ . include2.sh
I am being sourced, this filename is include2.sh and my caller script/shell name was bash
Do something with sourced script
所以这是毫无例外的工作,它没有使用脆弱的$_东西。这个技巧使用了BASH的自省功能,即内置变量FUNCNAME和BASH_SOURCE;请参阅bash手册页中的文档。
只有两个警告:
1)对am_I_called的调用必须发生在源脚本中,而不是在任何函数中,以免${FUNCNAME[1]}返回其他东西。是的…您本可以检查${FUNCNAME[2]},但这样做只会使您的工作更加困难。
2)函数am_I_called必须驻留在源脚本中,如果你想知道被包含的文件的名称。