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

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。

其他回答

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。

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。

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 ..."

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。