将数字转换为 bash 中的字符串枚举
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Convert numbers to enumeration of strings in bash
提问by User1
Using bash, I have a list of strings that I want to use to replace an int. Here's an example:
使用 bash，我有一个字符串列表，我想用它来替换 int。下面是一个例子：
day1=Monday day2=Tuesday day3=Wednesday day4=Thursday day5=Friday day6=Saturday day7=Sunday
If I have an int, $dow, to represent the day of the week, how do I print the actual string? I tried this:
如果我有一个 int $dow 来表示星期几，我该如何打印实际的字符串？我试过这个：
echo ${day`echo $dow`}
but get error of "bad substitution". How do I make this work? Note: I can change the $day variables to a list or something.
但得到“坏替代”的错误。我如何使这项工作？注意：我可以将 $day 变量更改为列表或其他内容。
采纳答案by Ignacio Vazquez-Abrams
Yes, it would be easiest to do this as an array:
是的，最简单的方法是将其作为数组执行：
day=([1]=Monday [2]=Tuesday ...)
echo "${day[dow]}"
回答by Pointy
case $dow in
[1234567]) eval echo '$day'$dow ;;
esac
I didn't want to get yelled at for an unsafe use of "eval" :-)
我不想因为不安全地使用“eval”而受到大吼:-)
There's probably a more "modern" way of doing this.
可能有一种更“现代”的方式来做到这一点。
回答by Jürgen H?tzel
You could use variable indirection:
您可以使用变量间接：
day_var=day$dow
echo ${!day_var}
But you should consider the use of arrays:
但是你应该考虑使用数组：
days=("Monday" "Tuesday" "Wednesday" "Thursday" "Friday" "Saturday" "Sunday")
echo ${days[$[$dow-1]]}
回答by user unknown
In June 1970 (you remember?) the month started with Monday:
1970 年 6 月（你还记得吗？）这个月从星期一开始：
for d in {1..7}
do
date -d 1970-06-0$d +%A
done
