带有列输出的 Bash 菜单

Question

我正在开发一个新项目，并使用了我以前使用过的标准 bash 菜单脚本。

我非常确定我可能没有在搜索引擎（Bash 菜单列输出）中使用正确的术语，并且从经验中剥夺了自己的经验，自己去发现它。我希望我能在这件事上得到一些帮助。

我的问题是我可以在哪里制作，以便菜单的输出（很长并且会更长）可以显示在 5 列中，而不是现在出现的长列表。也许是我自己处理菜单的方式？菜单中使用了太多 printf？

main_menu(){
    printf "    Main Menu\n"
    printf "\n"
    printf "    1) Option 1\n"
    printf "    2) Option 2\n"
    printf "    3) Option 3\n"
    printf "    4) Option 4\n"
    printf "    5) Option 5\n"
    printf "    6) Option 6\n"
    printf "    7) Option 7\n"
    printf "    8) Option 8\n"
    printf "    9) Option 9\n"
    printf "    10) Option 10\n"
    printf "    11) Option 11\n"
    printf "    12) Option 12\n"
    printf "    13) Option 13\n"
    printf "    14) Option 14\n"
    printf "    15) Option 15\n"
    printf "    16) Option 16\n"
    printf "    17) Option 17\n"
    printf "    18) Option 18\n"
    printf "    19) Option 19\n"
    printf "    20) Option 20\n"
    printf "    21) Option 21\n"
    printf "    22) Option 22\n"
    printf "    23) Option 23\n"
    printf "    24) Option 24\n"
    printf "\n"
    printf "    x) x to exit\n"
    printf "\n"
    printf "    Your choice: "
    read main

clear
while [[ "$main" != '\n' ]];
    do
    if [[ "$main" = "\n" ]]; then
      exit;
    else
      case $main in
    1) clear;
       opt_1;
        ;;
    2) clear;
       opt_2;
        ;;
    3) clear;
       opt_3;
        ;;
    4) clear;
       opt_2;
        ;;

<SNIP> This to avoid a stupidly long post as we are aware how the script works :P</SNIP>

        ;;
    x) clear;
      exit;
      ;;
      \n) clear;
      exit;
      ;;
    *) clear;
      option_picked "Pick an option from the menu";
      main_menu;
      ;;
      esac
    fi
  done
}
option_picked(){
    message=${@:-"Error: No message passed"}
    printf "${message}\n"
}

<snip> Below here the submenu system with the commands and other stuff to do</snip>

我说的输出是这样的：

    Main Menu

1) Option 1
2) option 2
3) option 3
4) option 4
   snip to keep it short
x) x to exit

Your choice: 

但我正在尝试的是类似于

Main Menu

1) Option 1      2) Option 2      3) option 3    4) option 4    5) Option 5
6) Option 6      7) Option 7      8) option 8    9) option 9    10) Option 10

Answer 1

使用pr 可能会做你想做的事。

#!/usr/bin/env bash

COLUMNS=$(tput cols)

main_menu(){
    printf "    Main Menu\n"
    printf "\n"
  {
    printf "    1) Option 1\n"
    printf "    2) Option 2\n"
    printf "    3) Option 3\n"
    printf "    4) Option 4\n"
    printf "    5) Option 5\n"
    printf "    6) Option 6\n"
    printf "    7) Option 7\n"
    printf "    8) Option 8\n"
    printf "    9) Option 9\n"
    printf "    10) Option 10\n"
    printf "    11) Option 11\n"
    printf "    12) Option 12\n"
    printf "    13) Option 13\n"
    printf "    14) Option 14\n"
    printf "    15) Option 15\n"
    printf "    16) Option 16\n"
    printf "    17) Option 17\n"
    printf "    18) Option 18\n"
    printf "    19) Option 19\n"
    printf "    20) Option 20\n"
    printf "    21) Option 21\n"
    printf "    22) Option 22\n"
    printf "    23) Option 23\n"
    printf "    24) Option 24\n"
    } | pr -t8 -w "$COLUMNS"
    printf "\n"
    printf "    x) x to exit\n"
    printf "    Your choice: "
    read main

clear
while [[ "$main" != '\n' ]]; do
    if [[ "$main" = "\n" ]]; then
      exit;
    else
      case $main in
    1) clear;
       opt_1;
        ;;
    2) clear;
       opt_2;
        ;;
    3) clear;
       opt_3;
        ;;
    4) clear;
       opt_2;
        ;;
    x) clear;
      exit;
      ;;
      \\n) clear;
      exit;
      ;;
    *) clear;
      option_picked "Pick an option from the menu";
      main_menu;
      ;;
      esac
    fi
  done
}
option_picked(){
  message=(${@:-"Error: No message passed"})
    printf '%s\n' "${message[*]}"
}

main_menu