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。

@mklement0答案的一个小补充。这是我在脚本中使用的自定义函数，用于确定它是否被引用:

replace_shell(){
   if [ -n "$ZSH_EVAL_CONTEXT" ]; then 
      case $ZSH_EVAL_CONTEXT in *:file*) echo "Zsh is sourced";; esac
   else
      case ${0##*/} in sh|dash|bash) echo "Bash is sourced";; esac
   fi
}

在函数中，zsh的“$ZSH_EVAL_CONTEXT”的输出是toplevel:file:shfunc，而不仅仅是toplevel:file;因此，*:file*应该解决这个问题。

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

我需要一个在[mac, linux]上使用bash的一行程序。版本>= 3，这些答案都不符合要求。

[[ ${BASH_SOURCE[0]} = $0 ]] && main "$@"

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。

我最终检查[[$_ == "$(type -p "$0")"]]

if [[ $_ == "$(type -p "$0")" ]]; then
    echo I am invoked from a sub shell
else
    echo I am invoked from a source command
fi

当使用curl时…| bash -s——ARGS运行远程脚本时，$0将只是bash而不是正常的/bin/bash运行实际的脚本文件，所以我使用-p“$0”来显示bash的完整路径。

测试:

curl -sSL https://github.com/jjqq2013/bash-scripts/raw/master/common/relpath | bash -s -- /a/b/c/d/e /a/b/CC/DD/EE

source <(curl -sSL https://github.com/jjqq2013/bash-scripts/raw/master/common/relpath)
relpath /a/b/c/d/e /a/b/CC/DD/EE

wget https://github.com/jjqq2013/bash-scripts/raw/master/common/relpath
chmod +x relpath
./relpath /a/b/c/d/e /a/b/CC/DD/EE