为什么fish中的or操作符似乎不起作用

Question

function shapeshift
   argparse 'h/help' -- $argv
   if not set -q argv[1] $argv or if set -q _flag_help 
      echo "Shapeshift lets you change into another root system."
      echo "It is a simple tool that mounts, and chroots a partition of your choice."
      echo ""
      echo "Use it in the following way:"
      set_color green; echo "shapeshift nvme0n1p4"; set_color normal
      echo ""
      echo "Here is a list of the available partitions:"
      echo ""
      sudo lsblk -o NAME,SIZE,MOUNTPOINTS,TYPE | awk '$4 == "part"'
   end
end

我似乎无法同时获得两者，用参数调用和用 --help 调用，似乎同时开始工作。

我尝试了这段代码的几个不同版本，一旦我用 or 将两个事件链接在一起，似乎总是会破坏代码。无论我先调用哪一个，或者是否在第二个命令前面使用单独的 if 语句。似乎总是失败。

Answer 1

我不知道为什么，但这似乎有效：

function shapeshift
   argparse 'h/help' -- $argv
   # Print the available partitions if no one given
   if not parameter_is_provided $argv; if set -q _flag_help
      echo "Shapeshift lets you change into another root system."
      echo "It is a simple tool that mounts, and chroots a partition of your choice."
      echo ""
      echo "Use it in the following way:"
      set_color green; echo "shapeshift nvme0n1p4"; set_color normal
      echo ""
      echo "Here is a list of the available partitions:"
      echo ""
      sudo lsblk -o NAME,SIZE,MOUNTPOINTS,TYPE | awk '$4 == "part"'
      return 0
   end
end