这并不完全是OP想要的，但我经常发现自己需要源代码脚本只是为了加载它的函数(即作为一个库)。例如，用于基准测试或测试目的。

下面是一个适用于所有shell(包括POSIX)的设计:

将所有顶级操作包装在run_main()函数中。 让您的源脚本检查不执行任何操作的初始-no-run参数;没有——no-run，它可以调用run_main。 使用以下方法获取脚本的源代码:

set -- --no-run "$@"
. script.sh
shift

问题是。或者来源是不可能将参数可移植地传递给脚本。POSIX shell忽略参数。并传递调用者的“$@”。

我认为在ksh和bash中没有任何可移植的方法来做到这一点。在bash中，您可以使用调用器输出来检测它，但我认为ksh中不存在等效的输出。

$_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。

我想对丹尼斯非常有用的回答提出一个小小的更正，让它更容易携带，我希望:

[ "$_" != "$0" ] && echo "Script is being sourced" || echo "Script is a subshell"

因为[[不被Debian POSIX兼容外壳识别(有些保留的IMHO)， dash。同样，在shell中，可能需要使用引号来防止文件名中包含空格。

FWIW，在阅读了所有其他的答案后，我想出了以下解决方案:

更新:事实上，有人在另一个答案中发现了一个已经更正的错误，这也影响了我的答案。我认为这里的更新也是一个改进(如果你好奇，请参阅编辑)。

这适用于所有以#!开头的脚本。/bin/bash，但也可以由不同的shell来获取一些信息(如设置)，这些信息保存在主函数之外。

根据下面的评论，这里的答案显然不适用于所有bash变体。同样不适用于/bin/sh基于bash的系统。也就是说，它失败的bash v3。在MacOS上。(目前我不知道如何解决这个问题。)

#!/bin/bash

# Function definitions (API) and shell variables (constants) go here
# (This is what might be interesting for other shells, too.)

# this main() function is only meant to be meaningful for bash
main()
{
# The script's execution part goes here
}

BASH_SOURCE=".$0" # cannot be changed in bash
test ".$0" != ".$BASH_SOURCE" || main "$@"

你可以使用以下(在我看来可读性较差)代码来代替最后两行，在其他shell中不设置BASH_SOURCE，并允许set -e在main中工作:

if ( BASH_SOURCE=".$0" && exec test ".$0" != ".$BASH_SOURCE" ); then :; else main "$@"; fi

这个脚本配方有以下属性:

If executed by bash the normal way, main is called. Please note that this does not include a call like bash -x script (where script does not contain a path), see below. If sourced by bash, main is only called, if the calling script happens to have the same name. (For example, if it sources itself or via bash -c 'someotherscript "$@"' main-script args.. where main-script must be, what test sees as $BASH_SOURCE). If sourced/executed/read/evaled by a shell other than bash, main is not called (BASH_SOURCE is always different to $0). main is not called if bash reads the script from stdin, unless you set $0 to be the empty string like so: ( exec -a '' /bin/bash ) <script If evaluated by bash with eval (eval "`cat script`" all quotes are important!) from within some other script, this calls main. If eval is run from commandline directly, this is similar to previous case, where the script is read from stdin. (BASH_SOURCE is blank, while $0 usually is /bin/bash if not forced to something completely different.) If main is not called, it does return true ($?=0). This does not rely on unexpected behavior (previously I wrote undocumented, but I found no documentation that you cannot unset nor alter BASH_SOURCE either): BASH_SOURCE is a bash reserved array. But allowing BASH_SOURCE=".$0" to change it would open a very dangerous can of worms, so my expectation is, that this must have no effect (except, perhaps, some ugly warning shows up in some future version of bash). There is no documentation that BASH_SOURCE works outside functions. However the opposite (that it only works in functions) is neither documented. The observation is, that it works (tested with bash v4.3 and v4.4, unfortunately I have no bash v3.x anymore) and that quite too many scripts would break, if $BASH_SOURCE stops working as observed. Hence my expectation is, that BASH_SOURCE stays as is for future versions of bash, too. In contrast (nice find, BTW!) consider ( return 0 ), which gives 0 if sourced and 1 if not sourced. This comes a bit unexpected not only for me , and (according to the readings there) POSIX says, that return from subshell is undefined behavior (and the return here is clearly from a subshell). Perhaps this feature eventually gets enough widespread use such that it can no more be changed, but AFAICS there is a much higher chance that some future bash version accidental changes the return behavior in that case. Unfortunately bash -x script 1 2 3 does not run main. (Compare script 1 2 3 where script has no path). Following can be used as workaround: bash -x "`which script`" 1 2 3 bash -xc '. script' "`which script`" 1 2 3 That bash script 1 2 3 does not run main can be considered a feature. Note that ( exec -a none script ) calls main (bash does not pass it's $0 to the script, for this you need to use -c as shown in the last point).

因此，除了一些极端情况外，main只在脚本以通常的方式执行时才被调用。通常这就是你想要的，特别是因为它缺乏复杂的难以理解的代码。

注意，它与Python代码非常相似: 如果__name__ == '__main__': main() 这也阻止了main的调用，除了一些角落的情况，如 你可以导入/加载脚本并强制__name__='__main__'

为什么我认为这是一个解决挑战的好方法

如果您有一些可以由多个shell提供源代码的东西，那么它必须是兼容的。然而(请阅读其他答案)，由于没有(容易实现的)可移植的方法来检测来源，您应该更改规则。

通过强制脚本必须由/bin/bash执行，您可以做到这一点。

这解决了所有的情况，但在以下情况下，脚本不能直接运行:

/bin/bash未安装或无法运行(即在引导环境中) 如果你将它管道到一个shell，比如curl https://example.com/script | $ shell (注意:这只适用于最近的bash。据报道，这种配方对某些变体无效。所以一定要检查它是否适用于你的情况。)

然而，我想不出任何真正的原因，在哪里你需要它，以及能力来源完全相同的脚本并行!通常，您可以将其包装起来手动执行main。像这样:

$SHELL -c '。脚本&& main {curl https://example.com/script && echo && echo main;} | $ shell $SHELL -c“eval”“curl https://example.com/script”“&& main” echo 'eval ' ' curl https://example.com/script ' ' && main' | $SHELL

笔记

如果没有其他答案的帮助，这个答案是不可能的!甚至是错误的——最初促使我发表这篇文章的原因。 更新:由于在https://stackoverflow.com/a/28776166/490291中发现的新发现而编辑

直截了当:您必须计算变量“$0”是否等于Shell的名称。

是这样的:

#!/bin/bash

echo "First Parameter: $0"
echo
if [[ "$0" == "bash" ]] ; then
    echo "The script was sourced."
else
    echo "The script WAS NOT sourced."
fi

通过壳:

$ bash check_source.sh 
First Parameter: check_source.sh

The script WAS NOT sourced.

通过来源:

$ source check_source.sh
First Parameter: bash

The script was sourced.

很难有一种100%可移植的方法来检测脚本是否来自源代码。

根据我的经验(使用Shellscripting 7年)，唯一安全的方法(不依赖具有pid等的环境变量，这是不安全的，因为它是变量)，你应该:

扩展你的“如果”的可能性 使用开关/箱子，如果你想。

这两个选项都不能自动缩放，但这是更安全的方式。

例如:

当您通过SSH会话源脚本时，变量“$0”(当使用source时)返回的值是-bash。

#!/bin/bash

echo "First Parameter: $0"
echo
if [[ "$0" == "bash" || "$0" == "-bash" ]] ; then
    echo "The script was sourced."
else
    echo "The script WAS NOT sourced."
fi

OR

#!/bin/bash

echo "First Parameter: $0"
echo
if [[ "$0" == "bash" ]] ; then
    echo "The script was sourced."
elif [[ "$0" == "-bash" ]] ; then
    echo "The script was sourced via SSH session."
else
    echo "The script WAS NOT sourced."
fi