Bash 脚本侦听按键以继续前进
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Bash script listen for key press to move on
提问by Cripto
So, I want to write a bash script that are a sequence of steps and ill identify it as "task#". However, each step is only completed and can run as long as the user wants.
所以，我想编写一个 bash 脚本，它是一系列步骤，并将其标识为“任务#”。但是，每个步骤都只是完成，并且可以根据用户的需要运行。
Do task1
if keypressed stop task1 and move on #this is the part I need help with. There can be up to 10 of these move on steps.
Do task2
...
kina like top; it keeps doing stuff until you hit q to quite, however, i want to move on to the next thing
基那像顶部; 它一直在做一些事情，直到你按 q 完全，但是，我想继续做下一件事
回答by kev
you can use readbuiltin command with option -tand -n
您可以使用read带有选项-t和的内置命令-n
while :
do
# TASK 1
date
read -t 1 -n 1 key
if [[ $key = q ]]
then
break
fi
done
# TASK 2
date +%s
回答by mklement0
kev's great solutionworks well even in Bash 3.x., but it introduces a 1-second delay(-t 1) in every loop iteration.
kev 的出色解决方案即使在 Bash 3.x 中也能很好地工作。，但它会在每次循环迭代中引入1 秒延迟( -t 1)。
In Bash 3.x, the lowest supported value for -t(timeout) is 1(second), unfortunately.
不幸的是，-t1在Bash 3.x 中，(timeout)支持的最低值是(second)。
Bash 4.x supports 0and fractionalvalues, however:
Bash 4.x 支持0和小数值，但是：
A solution that supports an arbitrary keysuch as qrequires a nonzero-tvalue, but you can specify a value very close to 0to minimize the delay:
支持任意键（例如q需要非零-t值）的解决方案，但您可以指定一个非常接近的值0以最小化延迟：
#!/bin/bash
# !! BASH 4.x+ ONLY
while :; do
# Loop command
date
# Check for 'q' keypress *waiting very briefly* and exit the loop, if found.
read -t 0.01 -r -s -N 1 && [[ $REPLY == 'q' ]] && break
done
# Post-loop command
date +%s
Caveat: The above uses 0.01as the almost-no-timeout value, but, depending on your host platform, terminal program and possibly CPU speed/configuration, a larger value may be required / a smaller value may be supported. If the value is too small, you'll see intermittent error setting terminal attributes: Interrupted system callerrors- if anyone knows why, do tell us.
警告：以上0.01用作几乎无超时值，但是，根据您的主机平台、终端程序和可能的 CPU 速度/配置，可能需要更大的值/可能支持更小的值。如果值太小，您会看到间歇性error setting terminal attributes: Interrupted system call错误- 如果有人知道原因，请告诉我们。
Tip of the hat to jarnofor his help with the following:
在帽子的提示亚诺他与以下帮助：
Using -t 0, works as follows, according to help read(emphasis added):
使用-t 0, 工作如下，根据help read（强调）：
If TIMEOUT is 0, read returns immediately, without trying to read any data, returning success only if input is availableon the specified file descriptor.
如果 TIMEOUT 为 0，则 read立即返回，不尝试读取任何数据，仅当指定文件描述符上的输入可用时才返回成功。
As of Bash v4.4.12 and 5.0.11, unfortunately, -t 0seems to ignore-n/ -N, so that only an ENTERkeypress (or a sequence of keypresses ending inENTER) causes readto indicate that data is available.
If anyone knows whether this is a bugor whether there's a good reason for this behavior, do let us know.
不幸的是，从 Bash v4.4.12 和 5.0.11 开始，-t 0似乎忽略了-n/ -N，因此只有ENTER按键（或以结尾的按键序列ENTER）导致read表明数据可用。
如果有人知道这是一个错误还是这种行为是否有充分的理由，请告诉我们。
Therefore, only with ENTERas the quit key is a -t 0solution currently possible:
因此，只有使用ENTER退出键才是-t 0目前可能的解决方案：
#!/bin/bash
# !! BASH 4.x+ ONLY
while :; do
# Loop command
date
# Check for ENTER keypress and, after clearing the input buffer
# with a dummy `read`, exit the loop.
read -t 0 -r -N 1 && { read -r; break; }
done
# Post-loop command
date +%s
