bash, ksh, zsh的健壮解决方案，包括一个跨shell解决方案，以及一个相当健壮的posix兼容解决方案:

给出的版本号是验证功能的版本号——很可能，这些解决方案也适用于更早的版本——欢迎反馈。 仅使用POSIX特性(例如在dash中，它在Ubuntu中充当/bin/sh)，没有可靠的方法来确定脚本是否被引用-请参阅下面的最佳近似方法。

重要的是:

The solutions determine whether the script is being sourced by its caller, which may be a shell itself or another script (which may or may not be sourced itself): Also detecting the latter case adds complexity; if you do not need to detect the case when your script is being sourced by another script, you can use the following, relatively simple POSIX-compliant solution: # Helper function is_sourced() { if [ -n "$ZSH_VERSION" ]; then case $ZSH_EVAL_CONTEXT in *:file:*) return 0;; esac else # Add additional POSIX-compatible shell names here, if needed. case ${0##*/} in dash|-dash|bash|-bash|ksh|-ksh|sh|-sh) return 0;; esac fi return 1 # NOT sourced. } # Sample call. is_sourced && sourced=1 || sourced=0 All solutions below must run in the top-level scope of your script, not inside functions.

下面是一行程序——解释如下;跨shell版本是复杂的，但它应该可以健壮地工作:

Bash(在3.57、4.4.19和5.1.16上验证)

(return 0 2>/dev/null) && sourced=1 || sourced=0

KSH(在93u+上验证)

[[ "$(cd -- "$(dirname -- "$0")" && pwd -P)/$(basename -- "$0")" != "$(cd -- "$(dirname -- "${.sh.file}")" && pwd -P)/$(basename -- "${.sh.file}")" ]] && sourced=1 || sourced=0

ZSH(5.0.5验证)

[[ $ZSH_EVAL_CONTEXT =~ :file$ ]] && sourced=1 || sourced=0

交叉shell (bash, ksh, zsh)

(
  [[ -n $ZSH_VERSION && $ZSH_EVAL_CONTEXT =~ :file$ ]] || 
  [[ -n $KSH_VERSION && "$(cd -- "$(dirname -- "$0")" && pwd -P)/$(basename -- "$0")" != "$(cd -- "$(dirname -- "${.sh.file}")" && pwd -P)/$(basename -- "${.sh.file}")" ]] || 
  [[ -n $BASH_VERSION ]] && (return 0 2>/dev/null)
) && sourced=1 || sourced=0

posix兼容;由于技术原因，不是一行程序(单一管道)，并且不完全健壮(见底部):

sourced=0
if [ -n "$ZSH_VERSION" ]; then 
  case $ZSH_EVAL_CONTEXT in *:file) sourced=1;; esac
elif [ -n "$KSH_VERSION" ]; then
  [ "$(cd -- "$(dirname -- "$0")" && pwd -P)/$(basename -- "$0")" != "$(cd -- "$(dirname -- "${.sh.file}")" && pwd -P)/$(basename -- "${.sh.file}")" ] && sourced=1
elif [ -n "$BASH_VERSION" ]; then
  (return 0 2>/dev/null) && sourced=1 
else # All other shells: examine $0 for known shell binary filenames.
     # Detects `sh` and `dash`; add additional shell filenames as needed.
  case ${0##*/} in sh|-sh|dash|-dash) sourced=1;; esac
fi

---

解释

---

bash

(return 0 2>/dev/null) && sourced=1 || sourced=0

注意:该技术改编自user5754163的答案，因为它比原来的解决方案更健壮，[[$0 != "$BASH_SOURCE"]] && sourced=1 || sourced=0[1]

Bash allows return statements only from functions and, in a script's top-level scope, only if the script is sourced. If return is used in the top-level scope of a non-sourced script, an error message is emitted, and the exit code is set to 1. (return 0 2>/dev/null) executes return in a subshell and suppresses the error message; afterwards the exit code indicates whether the script was sourced (0) or not (1), which is used with the && and || operators to set the sourced variable accordingly. Use of a subshell is necessary, because executing return in the top-level scope of a sourced script would exit the script. Tip of the hat to @Haozhun, who made the command more robust by explicitly using 0 as the return operand; he notes: per bash help of return [N]: "If N is omitted, the return status is that of the last command." As a result, the earlier version [which used just return, without an operand] produces incorrect result if the last command on the user's shell has a non-zero return value.

---

ksh

[[ "$(cd -- "$(dirname -- "$0")" && pwd -P)/$(basename -- "$0")" != "$(cd -- "$(dirname -- "${.sh.file}")" && pwd -P)/$(basename -- "${.sh.file}")" ]] && sourced=1 || sourced=0

特殊变量${.sh。file}有点类似于$BASH_SOURCE;注意${.sh。File}会在bash、zsh和dash中导致语法错误，所以一定要在多shell脚本中有条件地执行它。

与bash不同，$0和${.sh。file}不能保证是相同的-在不同的时候，一个可能是相对路径或仅仅是文件名，而另一个可能是完整的文件名;因此，$0和${.sh。File}在比较之前必须解析为全路径。如果完整路径不同，则隐含了源。

---

zsh

[[ $ZSH_EVAL_CONTEXT =~ :file$) ]] && sourced=1 || sourced=0

$ZSH_EVAL_CONTEXT包含关于计算上下文:子字符串文件的信息，用:分隔，只有在脚本被引用时才会出现。

在源脚本的顶级作用域中，$ZSH_EVAL_CONTEXT以:file结尾，这就是这个测试的限制。在函数内部，:shfunc被追加到:file;在命令替换中追加:cmdsubst。

---

仅使用POSIX特性

如果您愿意做出某些假设，那么您可以根据可能正在执行脚本的shell的二进制文件名，对脚本是否被引用做出合理但并非万无一错的猜测。 值得注意的是，这意味着当您的脚本被另一个脚本引用时，这种方法不会检测到这种情况。

本回答中的“如何处理源调用”一节只详细讨论了POSIX特性无法处理的边缘情况。

检查二进制文件名依赖于$0的标准行为，例如，zsh就没有这种行为。

因此，最安全的方法是将上述健壮的、特定于shell的方法(不依赖于$0)与所有剩余shell的基于$0的回退解决方案结合起来。

简而言之:解决方案如下:

在包含特定于shell的测试的shell中:稳定地工作。 在所有其他shell中:仅当脚本直接来自这样的shell(而不是来自另一个脚本)时才能正常工作。

向Stéphane Desneux和他的灵感答案致敬(将我的跨shell语句表达式转换为sh兼容的if语句，并为其他shell添加一个处理程序)。

sourced=0
if [ -n "$ZSH_VERSION" ]; then 
  case $ZSH_EVAL_CONTEXT in *:file) sourced=1;; esac
elif [ -n "$KSH_VERSION" ]; then
  [ "$(cd -- "$(dirname -- "$0")" && pwd -P)/$(basename -- "$0")" != "$(cd -- "$(dirname -- "${.sh.file}")" && pwd -P)/$(basename -- "${.sh.file}")" ] && sourced=1
elif [ -n "$BASH_VERSION" ]; then
  (return 0 2>/dev/null) && sourced=1 
else # All other shells: examine $0 for known shell binary filenames.
  # Detects `sh` and `dash`; add additional shell filenames as needed.
  case ${0##*/} in sh|-sh|dash|-dash) sourced=1;; esac
fi

注意，为了健壮性，每个shell二进制文件名(例如sh)都表示了两次——一次是原样，第二次是，前缀是-。这是为了考虑诸如macOS这样的环境，其中交互式shell作为登录shell启动，其自定义值为$0，该值是前缀为-的(无路径)shell文件名。谢谢,t7e。 (虽然sh和dash可能不太可能用作交互式shell，但您可能需要将其他shell添加到列表中。)

---

[1] user1902689发现[[$0 != "$BASH_SOURCE"]]在通过将其文件名传递给bash二进制文件执行位于$PATH中的脚本时，会产生假阳性;例如，bash my-script，因为$0只是my-script，而$BASH_SOURCE是完整路径。虽然您通常不会使用这种技术来调用$PATH中的脚本—您只会直接调用它们(my-script)—当与-x结合使用时，它对调试很有帮助。

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。

这在后面的脚本中起作用，不依赖于_变量:

## Check to make sure it is not sourced:
Prog=myscript.sh
if [ $(basename $0) = $Prog ]; then
   exit 1  # not sourced
fi

or

[ $(basename $0) = $Prog ] && exit

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

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。