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
可以在POSIX shell中处理区域设置感知的“是/否选项”;通过使用LC_MESSAGES语言环境类别的条目,witch提供了与输入匹配的现成RegEx模式,以及本地化Yes No的字符串。
#!/usr/bin/env sh
# Getting LC_MESSAGES values into variables
# shellcheck disable=SC2046 # Intended IFS splitting
IFS='
' set -- $(locale LC_MESSAGES)
yesexpr="$1"
noexpr="$2"
yesstr="$3"
nostr="$4"
messages_codeset="$5" # unused here, but kept as documentation
# Display Yes / No ? prompt into locale
echo "$yesstr / $nostr ?"
# Read answer
read -r yn
# Test answer
case "$yn" in
# match only work with the character class from the expression
${yesexpr##^}) echo "answer $yesstr" ;;
${noexpr##^}) echo "answer $nostr" ;;
esac
编辑:正如@Urhixidur在评论中提到的:
不幸的是,POSIX只指定了前两个(yeexpr和noexpr)。在Ubuntu 16上,yestr和nostr为空。
参见:https://www.ee.ryerson.ca/~courses/ele709/susv4/xrat/V_4_xbd_chap07.html#tag_21_07_03_06
LC_消息yestr和nostr语言环境关键字以及yesstr和nostr语言信息项以前用于匹配用户的肯定和否定响应。在POSIX.1-2008中,yeexpr、noexpr、yesexpr和noexpr扩展正则表达式已取代它们。应用程序应该使用通用的基于语言环境的消息传递工具来发出包含所需响应示例的提示消息。
或者使用Bash方式的区域设置:
#!/usr/bin/env bash
IFS=$'\n' read -r -d '' yesexpr noexpr _ < <(locale LC_MESSAGES)
printf -v yes_or_no_regex "(%s)|(%s)" "$yesexpr" "$noexpr"
printf -v prompt $"Please answer Yes (%s) or No (%s): " "$yesexpr" "$noexpr"
declare -- answer=;
until [[ "$answer" =~ $yes_or_no_regex ]]; do
read -rp "$prompt" answer
done
if [[ -n "${BASH_REMATCH[1]}" ]]; then
echo $"You answered: Yes"
else
echo $"No, was your answer."
fi
使用语言环境提供的regexp来匹配答案。
要翻译剩余的消息,请使用bash--dump po string scriptname输出po字符串以进行本地化:
#: scriptname:8
msgid "Please answer Yes (%s) or No (%s): "
msgstr ""
#: scriptname:17
msgid "You answered: Yes"
msgstr ""
#: scriptname:19
msgid "No, was your answer."
msgstr ""
其他回答
仅单个按键
这里有一个更长但可重复使用的模块化方法:
返回0=yes和1=no无需按enter键-仅需一个字符可以按enter键接受默认选择可以禁用默认选项以强制选择适用于zsh和bash。
按enter键时默认为“否”
请注意,N是大写的。此处按enter键,接受默认值:
$ confirm "Show dangerous command" && echo "rm *"
Show dangerous command [y/N]?
还要注意,[y/N]?已自动追加。默认的“no”被接受,因此不会有任何回音。
重新提示,直到给出有效响应:
$ confirm "Show dangerous command" && echo "rm *"
Show dangerous command [y/N]? X
Show dangerous command [y/N]? y
rm *
按回车键时默认为“是”
请注意,Y是大写的:
$ confirm_yes "Show dangerous command" && echo "rm *"
Show dangerous command [Y/n]?
rm *
上面,我刚刚按了回车键,所以命令运行了。
输入时无默认值-需要y或n
$ get_yes_keypress "Here you cannot press enter. Do you like this [y/n]? "
Here you cannot press enter. Do you like this [y/n]? k
Here you cannot press enter. Do you like this [y/n]?
Here you cannot press enter. Do you like this [y/n]? n
$ echo $?
1
这里,返回了1或false。请注意,使用此较低级别的函数,您需要提供自己的[y/n]?促使
Code
# Read a single char from /dev/tty, prompting with "$*"
# Note: pressing enter will return a null string. Perhaps a version terminated with X and then remove it in caller?
# See https://unix.stackexchange.com/a/367880/143394 for dealing with multi-byte, etc.
function get_keypress {
local REPLY IFS=
>/dev/tty printf '%s' "$*"
[[ $ZSH_VERSION ]] && read -rk1 # Use -u0 to read from STDIN
# See https://unix.stackexchange.com/q/383197/143394 regarding '\n' -> ''
[[ $BASH_VERSION ]] && </dev/tty read -rn1
printf '%s' "$REPLY"
}
# Get a y/n from the user, return yes=0, no=1 enter=$2
# Prompt using $1.
# If set, return $2 on pressing enter, useful for cancel or defualting
function get_yes_keypress {
local prompt="${1:-Are you sure [y/n]? }"
local enter_return=$2
local REPLY
# [[ ! $prompt ]] && prompt="[y/n]? "
while REPLY=$(get_keypress "$prompt"); do
[[ $REPLY ]] && printf '\n' # $REPLY blank if user presses enter
case "$REPLY" in
Y|y) return 0;;
N|n) return 1;;
'') [[ $enter_return ]] && return "$enter_return"
esac
done
}
# Credit: http://unix.stackexchange.com/a/14444/143394
# Prompt to confirm, defaulting to NO on <enter>
# Usage: confirm "Dangerous. Are you sure?" && rm *
function confirm {
local prompt="${*:-Are you sure} [y/N]? "
get_yes_keypress "$prompt" 1
}
# Prompt to confirm, defaulting to YES on <enter>
function confirm_yes {
local prompt="${*:-Are you sure} [Y/n]? "
get_yes_keypress "$prompt" 0
}
用最少的行数实现这一点的最简单方法如下:
read -p "<Your Friendly Message here> : y/n/cancel" CONDITION;
if [ "$CONDITION" == "y" ]; then
# do something here!
fi
if只是一个例子:如何处理这个变量取决于您。
在shell提示符下获取用户输入的最简单和最广泛的方法是read命令。说明其用途的最佳方法是简单的演示:
while true; do
read -p "Do you wish to install this program? " yn
case $yn in
[Yy]* ) make install; break;;
[Nn]* ) exit;;
* ) echo "Please answer yes or no.";;
esac
done
Steven Huwig指出的另一种方法是Bash的select命令。下面是使用select的相同示例:
echo "Do you wish to install this program?"
select yn in "Yes" "No"; do
case $yn in
Yes ) make install; break;;
No ) exit;;
esac
done
使用select,您不需要清理输入–它会显示可用的选项,并键入与您的选择相对应的数字。它也会自动循环,所以如果它们提供了无效的输入,就不需要使用while true循环来重试。
此外,Léa Gris在她的回答中展示了一种使请求语言不可知的方法。调整我的第一个示例以更好地服务于多种语言可能如下所示:
set -- $(locale LC_MESSAGES)
yesexpr="$1"; noexpr="$2"; yesword="$3"; noword="$4"
while true; do
read -p "Install (${yesword} / ${noword})? " yn
if [[ "$yn" =~ $yesexpr ]]; then make install; exit; fi
if [[ "$yn" =~ $noexpr ]]; then exit; fi
echo "Answer ${yesword} / ${noword}."
done
显然,其他通信字符串在这里仍然没有翻译(Install,Answer),这需要在更完整的翻译中解决,但在许多情况下,即使是部分翻译也会很有用。
最后,请查看F.Hauri的优秀答案。
更通用的是:
function menu(){
title="Question time"
prompt="Select:"
options=("Yes" "No" "Maybe")
echo "$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";;
$(( ${#options[@]}+1 )) ) clear; echo "Goodbye!"; exit;;
*) echo "Invalid option. Try another one.";continue;;
esac
done
return
}
在我的情况下,我需要从下载的脚本中读取。,
curl -Ss https://example.com/installer.sh | sh
在这种情况下,read-ryn</dev/tty行允许它读取输入。
printf "These files will be uploaded. Is this ok? (y/N) "
read -r yn </dev/tty
if [ "$yn" = "y" ]; then
# Yes
else
# No
fi