如何使用 sudo 執行 bash 函式？

我在全域 bashrc 中定義了一個 bash 函式，它需要 root 權限才能工作。如何使用 sudo 執行它，例如sudo myfunction. 預設情況下，它會給出錯誤：

sudo：myfunction：找不到命令

Luca 好心地向我指出了這個問題，這是我的方法：在呼叫 sudo 之前擴展函式/別名並將其全部傳遞給 sudo，不需要臨時文件。

在我的部落格上進行了解釋。有很多報價處理:-)

# Wrap sudo to handle aliases and functions
# Wout.Mertens@gmail.com
#
# Accepts -x as well as regular sudo options: this expands variables as you not root
#
# Comments and improvements welcome
#
# Installing: source this from your .bashrc and set alias sudo=sudowrap
#  You can also wrap it in a script that changes your terminal color, like so:
#  function setclr() {
#   local t=0               
#   SetTerminalStyle $1                
#   shift
#   "$@"
#   t=$?
#   SetTerminalStyle default
#   return $t
#  }
#  alias sudo="setclr sudo sudowrap"
#  If SetTerminalStyle is a program that interfaces with your terminal to set its
#  color.

# Note: This script only handles one layer of aliases/functions.

# If you prefer to call this function sudo, uncomment the following
# line which will make sure it can be called that
#typeset -f sudo >/dev/null && unset sudo

sudowrap () 
{
   local c="" t="" parse=""
   local -a opt
   #parse sudo args
   OPTIND=1
   i=0
   while getopts xVhlLvkKsHPSb:p:c:a:u: t; do
       if [ "$t" = x ]; then
           parse=true
       else
           opt[$i]="-$t"
           let i++
           if [ "$OPTARG" ]; then
               opt[$i]="$OPTARG"
               let i++
           fi
       fi
   done
   shift $(( $OPTIND - 1 ))
   if [ $# -ge 1 ]; then
       c="$1";
       shift;
       case $(type -t "$c") in 
       "")
           echo No such command "$c"
           return 127
           ;;
       alias)
           c="$(type "$c")"
           # Strip "... is aliased to `...'"
           c="${c#*\`}"
           c="${c%\'}"
           ;;
       function)
           c="$(type "$c")"
           # Strip first line
           c="${c#* is a function}"
           c="$c;\"$c\""
           ;;
       *)
           c="\"$c\""
           ;;
       esac
       if [ -n "$parse" ]; then
           # Quote the rest once, so it gets processed by bash.
           # Done this way so variables can get expanded.
           while [ -n "$1" ]; do
               c="$c \"$1\""
               shift
           done
       else
           # Otherwise, quote the arguments. The echo gets an extra
           # space to prevent echo from parsing arguments like -n
           while [ -n "$1" ]; do
               t="${1//\'/\'\\\'\'}"
               c="$c '$t'"
               shift
           done
       fi
       echo sudo "${opt[@]}" -- bash -xvc \""$c"\" >&2
       command sudo "${opt[@]}" bash -xvc "$c"
   else
       echo sudo "${opt[@]}" >&2
       command sudo "${opt[@]}"
   fi
}
# Allow sudowrap to be used in subshells
export -f sudowrap

這種方法的一個缺點是它只擴展了您正在呼叫的函式，而不是您從那裡引用的任何額外函式。如果您引用載入在 bashrc 中的函式（假設它在bash -c呼叫時執行），Kyle 的方法可能會更好地處理這個問題。

引用自：https://serverfault.com/questions/177699