linux - Bash script run order, why does it queue the read command?

Question

I'm very confused at the moment.

What I want to do is create a listener function, that wait for two commands - and the thing is it should only listen every two hundred milliseconds and ignore the other input from the user.

function foo() {

    read -sN1 _input

    case "${_input}" in

        A) echo 'Option A';;
        B) echo 'Option B';;
    esac
}

while true; do
    sleep 0.2
    foo
done

If I "hammer" the A-key (or UP-key) ten times, it writes "Option A" ten times (although, slowly) - when it should only had time to write it at most three times. Why is that - and how do I fix it?

Answer 1

您的终端缓冲程序的输入。为了让你的程序忽略它在睡眠时收到的输入，你应该在调用之前清除输入缓冲区read。据我所知，在 bash 中没有办法做到这一点。

Answer 2

您可以将睡眠功能放在标准输入关闭的块中：

  {
    sleep 0.2
  } <&- 

或者，您可以在睡眠后立即清除（读取和丢弃）标准输入缓冲区：

sleep 0.2
read -t 1 -n 10000 discard

Answer 3

这就是我想出的有效方法。这不是很好，但它有效。

function inputCommand() {

    _now=`date +%s%N | cut -b1-13`
    _time=$(($_now-$_timeout))

    if [ $_time -gt 500 ]; then
            $1
            _timeout=`date +%s%N | cut -b1-13`
    fi
}

function keyPressUp() {

    echo "Key Press Up"
}

function waitForInput() {

    unset _input
    read -sN1 _input

    case "${_input}" in

            A) inputCommand "keyPressUp";;
    esac
}

while true; do

        waitForInput
done