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将每个输出存储到变量中

tag-icon • tag-icon bash variable for
将每个输出存储到变量中

我这里有一个脚本，它将列出用户输入的日期并输出 5 天前的日期。

#!/bin/bash
echo "What month?"
            echo "1 - January"
            echo "2 - February"
            echo "3 - March"
            echo "4 - April"
            echo "5 - May"
            echo "6 - June"
            echo "7 - July"
            echo "8 - August"
            echo "9 - September"
            echo "10 - October"
            echo "11 - November"
            echo "12 - December"
            echo ""
            echo -n "What month?  "
            read m

            if [ "$m" == "1" ]
            then
            mn="Jan "
            elif [ "$m" == "2" ]
            then
            mn="Feb "
            elif [ "$m" == "3" ]
            then
            mn="Mar "
            elif [ "$m" == "4" ]
            then
            mn="Apr "
            elif [ "$m" == "5" ]
            then
            mn="May "
            elif [ "$m" == "6" ]
            then
            mn="Jun "
            elif [ "$m" == "7" ]
            then
            mn="Jul "
            elif [ "$m" == "8" ]
            then
            mn="Aug "
            elif [ "$m" == "9" ]
            then
            mn="Sep "
            elif [ "$m" == "10" ]
            then
            mn="Oct "
            elif [ "$m" == "11" ]
            then
            mn="Nov "
            elif [ "$m" == "12" ]
            then
            mn="Dec "
            else
            echo "Invalid month"
            fi
            echo ""

            #DAY
            echo -n "What day? "
            read d
            if [ "$d" -lt "9" ]
            then
                    mnd="$mn"" ""$d"
            elif [ "$d"  -gt "31" ]
            then
                    mnd="1"
            else
                    mnd="$mn""$d"
            fi



 for dy in {0..4}; do
    date -d "$mn $d - $dy days" +'%b %_d'
 done

输出：

What month?
8
What day?
1

Aug  1
Jul 31
Jul 30
Jul 29
Jul 28

我现在想要的是将每个日期存储到变量中，例如第一行必须Aug 1存储在变量中x1 Jul 31必须存储在变量中x2等等。我的意思是第一个列表上的任何输出都必须存储在x1等等。

答案1

使用数组而不是尝试创建单独的变量。

declare -a x

for dy in {0..4}; do
  x+=( "$( date -d "$mn $d - $dy days" +'%b %_d' )" )
done

${x[0]}然后，您可以访问到中的四个值${x[3]}。

对于脚本的第一部分，您是否考虑过使用select语句？

select mn in "Jan" "Feb" "Mar" "Apr" "May" "Jun" \
             "Jul" "Aug" "Sep" "Oct" "Nov" "Dec"
do
  if [[ -z "$mn" ]]; then
    echo "Invalid choice" >&2
  else
    break
  fi
done

printf "You selected '%s'\n" "$mn"

这会执行以下操作：

1) Jan    3) Mar   5) May   7) Jul   9) Sep  11) Nov
2) Feb    4) Apr   6) Jun   8) Aug  10) Oct  12) Dec
#? 56
Invalid choice
#? 5
You selected 'May'

的值$mn将是选定的字符串（May例如）。

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