Bash 读取退格按钮行为问题

在 bash 中使用 read 时，按退格键不会删除最后输入的字符，但会在输入缓冲区中追加一个退格键。有什么方法可以更改它，以便删除删除从输入中键入的最后一个键？如果是这样怎么办？

这是一个简短的示例程序，我正在使用它，如果它有任何帮助的话：

#!/bin/bash

colour(){ #$1=text to colourise $2=colour id
        printf "%s%s%s" $(tput setaf $2) "$1" $(tput sgr0)
}
game_over() { #$1=message $2=score      
        printf "\n%s\n%s\n" "$(colour "Game Over!" 1)" "$1"
        printf "Your score: %s\n" "$(colour $2 3)"
        exit 0
}

score=0
clear
while true; do
        word=$(shuf -n1 /usr/share/dict/words) #random word from dictionary 
        word=${word,,} #to lower case
        len=${#word}
        let "timeout=(3+$len)/2"
        printf "%s  (time %s): " "$(colour $word 2)" "$(colour $timeout 3)"
        read -t $timeout -n $len input #read input here
        if [ $? -ne 0 ]; then   
                game_over "You did not answer in time" $score
        elif [ "$input" != "$word" ]; then
                game_over "You did not type the word correctly" $score;
        fi  
        printf "\n"
        let "score+=$timeout" 
done

选项-n nchars将终端转变为原始模式，所以你最好的机会是依靠readline (-e) [docs]:

$ read -n10 -e VAR

顺便说一句，好主意，尽管我会把单词的结尾留给用户（这是按下时的下意识反应）return).