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

read -e -p "Enter your choice: " choice

-e选项允许用户使用箭头键编辑输入。

如果要使用建议作为输入:

read -e -i "yes" -p "Enter your choice: " choice

-i选项打印提示性输入。

其他回答

多选版本:

ask () {                        # $1=question $2=options
    # set REPLY
    # options: x=..|y=..
    while $(true); do
        printf '%s [%s] ' "$1" "$2"
        stty cbreak
        REPLY=$(dd if=/dev/tty bs=1 count=1 2> /dev/null)
        stty -cbreak
        test "$REPLY" != "$(printf '\n')" && printf '\n'
        (
            IFS='|'
            for o in $2; do
                if [ "$REPLY" = "${o%%=*}" ]; then
                    printf '\n'
                    break
                fi
            done
        ) | grep ^ > /dev/null && return
    done
}

例子:

$ ask 'continue?' 'y=yes|n=no|m=maybe'
continue? [y=yes|n=no|m=maybe] g
continue? [y=yes|n=no|m=maybe] k
continue? [y=yes|n=no|m=maybe] y
$

它会将REPLY设置为y(在脚本中)。

您需要:

Bash内置命令(即便携式)检查TTY默认答案超时彩色问题

一小条

do_xxxx=y                      # In batch mode => Default is Yes
[[ -t 0 ]] &&                  # If TTY => Prompt the question
read -n 1 -p $'\e[1;32m
Do xxxx? (Y/n)\e[0m ' do_xxxx  # Store the answer in $do_xxxx
if [[ $do_xxxx =~ ^(y|Y|)$ ]]  # Do if 'y' or 'Y' or empty
then
    xxxx
fi

解释

[[-t 0]]&&read…=>如果TTY,调用命令读取read-n 1=>等待一个字符$'\e[1;32m…\e[0m'=>绿色打印(绿色很好,因为在白色/黑色背景上都可以阅读)[[$do_xxxx=~^(y|y|)$]]=>bash正则表达式

超时=>默认答案为否

do_xxxx=y
[[ -t 0 ]] && {                   # Timeout 5 seconds (read -t 5)
read -t 5 -n 1 -p $'\e[1;32m
Do xxxx? (Y/n)\e[0m ' do_xxxx ||  # read 'fails' on timeout
do_xxxx=n ; }                     # Timeout => answer No
if [[ $do_xxxx =~ ^(y|Y|)$ ]]
then
    xxxx
fi

检查此项

read -p "Continue? (y/n): " confirm && [[ $confirm == [yY] || $confirm == [yY][eE][sS] ]] || exit 1

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

这是我整理的东西:

#!/bin/sh

promptyn () {
    while true; do
        read -p "$1 " yn
        case $yn in
            [Yy]* ) return 0;;
            [Nn]* ) return 1;;
            * ) echo "Please answer yes or no.";;
        esac
    done
}

if promptyn "is the sky blue?"; then
    echo "yes"
else
    echo "no"
fi

我是一个初学者,所以要谨慎对待,但它似乎奏效了。