Bash：如何打印并运行其中包含管道运算符 | 的 cmd 数组

这是我的问题的后续：当命令的参数带有空格或要扩展的内容时，如何编写 bash 函数来打印和运行命令 https://stackoverflow.com/q/71117953/4561887

假设我有这个函数来打印并运行存储在数组中的命令：

# Print and run the cmd stored in the passed-in array
print_and_run() {
    echo "Running cmd:  $*"
    # run the command by calling all elements of the command array at once
    "$@"
}

这工作正常：

cmd_array=(ls -a /)
print_and_run "${cmd_array[@]}"

但这不起作用：

cmd_array=(ls -a / | grep "home")
print_and_run "${cmd_array[@]}"

Error: syntax error near unexpected token `|':

eRCaGuy_hello_world/bash$ ./print_and_run.sh 
./print_and_run.sh: line 55: syntax error near unexpected token `|'
./print_and_run.sh: line 55: `cmd_array=(ls -a / | grep "home")'

我怎样才能让这个概念与管道操作员一起工作（|）在命令中？

如果你想处理一个只包含|作为生成管道的指令，您可以这样做。我不推荐它 - 如果您不验证字符串中的变量不能仅包含单个管道字符，则意味着您存在安全风险 - 但这是可能的。

下面，我们创建一个随机的一次性"$pipe"使攻击变得更加困难的印记。如果你不愿意这样做，那就改变[[ $arg = "$pipe" ]] to [[ $arg = "|" ]].

# generate something random to make an attacker's job harder
pipe=$(uuidgen)

# use that randomly-generated sigil in place of | in our array
cmd_array=(
  ls -a /
  "$pipe" grep "home"
)

exec_array_pipe() {
  local arg cmd_q
  local -a cmd=( )
  while (( $# )); do
    arg=$1; shift
    if [[ $arg = "$pipe" ]]; then
      # log an eval-safe copy of what we're about to run
      printf -v cmd_q '%q ' "${cmd[@]}"
      echo "Starting pipeline component: $cmd_q" >&2
      # Recurse into a new copy of ourselves as a child process
      "${cmd[@]}" | exec_array_pipe "$@"
      return
    fi
    cmd+=( "$arg" )
  done
  printf -v cmd_q '%q ' "${cmd[@]}"
  echo "Starting pipeline component: $cmd_q" >&2
  "${cmd[@]}"
}

exec_array_pipe "${cmd_array[@]}"

查看它在在线沙箱中运行：https://ideone.com/IWOTfO https://ideone.com/IWOTfO