Question

我在bash中written up是一个'挂人'型游戏。目前正在工作，但我无法理解如何覆盖字符串中的某些字符。

所以目前我从用户那里收到一封信（例如'l'），并根据字符串'hello'进行检查，并打印出'_ _ ll _'（代码中{$prev）我想当用户输入'e'时，$prev更新为'_ ell _'等等等等等等'h'和'o'。

问：我如何更改字符串中的某个字符？

Answer 1

如果要操作变量中的单个字符，则更容易将所有字符保存在数组中。这样您就可以根据数组索引更改字符。下面显示了在操作字符时转换为数组或从数组转换的简单方法。注意：无需转换回变量，只需在数组中保持猜测的单词：

#!/bin/bash

declare -a array

prev="__||_"    ## previous state of guessing
printf "%s\n" "$prev"

for ((i=0; i<${#prev}; i++)); do   ## convert to array
    array+=( ${prev:i:1} )
done

array[1]="e"                        ## update letter

## convert back to string
printf -v prev $(echo ${array[@]} | tr -d ' ')
printf "%s\n" "$prev"

<强>输出

$ bash ~/scr/tmp/stack/strtoarray.sh
__||_
_e||_

Answer 2

使用数组作为单词和答案的hangman实现。它从$ {words}指向的文件中选择一个随机单词。

#!/usr/bin/env bash

words='/usr/share/dict/words'

die() { >&2 echo $@; exit 1; }

join() { local IFS="$1"; shift; echo "$*"; }

new_word() {
  unset word
  local length=$(wc -l < ${words})
  local count=$((${RANDOM} % ${length}))
  head -${count} ${words} | tail -1 | tr '[:upper:]' '[:lower:]'
}

draw_hangman() {
  # draw hangman here
  ((++guesses))
}

[[ -f ${words} ]] || die "Missing words file: ${words}"

declare -i guesses=0
declare -a word=( $(new_word | grep -o .) )
declare -a answer=( $(printf '_ %.0s' $(seq 1 ${#word[@]})) )

while :; do
  echo -e "\nPress - to quit.\n"
  echo -e "${answer[@]}\n"
  read -s -n 1 -p "$((${#word[@]} - ${guesses})) more guesses: " guess
  echo
  [[ ${guess} == '-' ]] && echo && break # quit
  [[ ${guesses} -eq $((${#word[@]} - 1)) ]] && echo -e "\nYOU LOSE!\n" && \
    echo "The word was $(join '' ${word[@]})" && break
  [[ ${word[@]} != *${guess}* ]] && draw_hangman && continue
  for i in ${!word[@]}; do
    [[ ${word[${i}]} == ${guess} ]] && answer[${i}]=${guess}
  done
  [[ ${answer[@]} == ${word[@]} ]] && echo -e "\nYOU WIN!\n" && break
done

如何更改/覆盖字符串中的某些字符？

2 个答案: