我想暂停shell脚本中的输入,并提示用户进行选择。标准的是、否或取消类型问题。如何在典型的bash提示符中完成此操作?


当前回答

是/否/取消

作用

#!/usr/bin/env bash
@confirm() {
  local message="$*"
  local result=''

  echo -n "> $message (Yes/No/Cancel) " >&2

  while [ -z "$result" ] ; do
    read -s -n 1 choice
    case "$choice" in
      y|Y ) result='Y' ;;
      n|N ) result='N' ;;
      c|C ) result='C' ;;
    esac
  done

  echo $result
}

用法

case $(@confirm 'Confirm?') in
  Y ) echo "Yes" ;;
  N ) echo "No" ;;
  C ) echo "Cancel" ;;
esac

使用干净的用户输入进行确认

作用

#!/usr/bin/env bash
@confirm() {
  local message="$*"
  local result=3

  echo -n "> $message (y/n) " >&2

  while [[ $result -gt 1 ]] ; do
    read -s -n 1 choice
    case "$choice" in
      y|Y ) result=0 ;;
      n|N ) result=1 ;;
    esac
  done

  return $result
}

用法

if @confirm 'Confirm?' ; then
  echo "Yes"
else
  echo "No"
fi

其他回答

更通用的是:

function menu(){
    title="Question time"
    prompt="Select:"
    options=("Yes" "No" "Maybe")
    echo "$title"
    PS3="$prompt"
    select opt in "${options[@]}" "Quit/Cancel"; do
        case "$REPLY" in
            1 ) echo "You picked $opt which is option $REPLY";;
            2 ) echo "You picked $opt which is option $REPLY";;
            3 ) echo "You picked $opt which is option $REPLY";;
            $(( ${#options[@]}+1 )) ) clear; echo "Goodbye!"; exit;;
            *) echo "Invalid option. Try another one.";continue;;
         esac
     done
     return
}

使用PyInquirer的一行python替代方案

python3 -c 'import PyInquirer; print(PyInquirer.prompt([{"type":"confirm", "message":"Do you want to continue?", "name":"r"}]).get("r"))'

它支持yes/no/cancel(intr,CTRL+C)。

是/否/取消

作用

#!/usr/bin/env bash
@confirm() {
  local message="$*"
  local result=''

  echo -n "> $message (Yes/No/Cancel) " >&2

  while [ -z "$result" ] ; do
    read -s -n 1 choice
    case "$choice" in
      y|Y ) result='Y' ;;
      n|N ) result='N' ;;
      c|C ) result='C' ;;
    esac
  done

  echo $result
}

用法

case $(@confirm 'Confirm?') in
  Y ) echo "Yes" ;;
  N ) echo "No" ;;
  C ) echo "Cancel" ;;
esac

使用干净的用户输入进行确认

作用

#!/usr/bin/env bash
@confirm() {
  local message="$*"
  local result=3

  echo -n "> $message (y/n) " >&2

  while [[ $result -gt 1 ]] ; do
    read -s -n 1 choice
    case "$choice" in
      y|Y ) result=0 ;;
      n|N ) result=1 ;;
    esac
  done

  return $result
}

用法

if @confirm 'Confirm?' ; then
  echo "Yes"
else
  echo "No"
fi

此解决方案读取单个字符,并在yes响应时调用函数。

read -p "Are you sure? (y/n) " -n 1
echo
if [[ $REPLY =~ ^[Yy]$ ]]; then
    do_something      
fi

我注意到,对于这样简单的用户输入,没有人发布显示多行回声菜单的答案,所以我的做法如下:

#!/bin/bash

function ask_user() {    

echo -e "
#~~~~~~~~~~~~#
| 1.) Yes    |
| 2.) No     |
| 3.) Quit   |
#~~~~~~~~~~~~#\n"

read -e -p "Select 1: " choice

if [ "$choice" == "1" ]; then

    do_something

elif [ "$choice" == "2" ]; then

    do_something_else

elif [ "$choice" == "3" ]; then

    clear && exit 0

else

    echo "Please select 1, 2, or 3." && sleep 3
    clear && ask_user

fi
}

ask_user

发布这种方法是希望有人发现它有用且省时。