Bash十进制到基数62的转换 [英] Bash decimal to base 62 conversion

问题描述

我想撤消以下 bash 命令执行的操作:

I would like to reverse the operation performed by the following bash command:

$ echo $((62#a39qrT))
9207903953

即将十进制9207903953 转换为 base 62 ，并保持 bash 的标准为 {0..9}，{a..z}，{A..Z} .

i.e. convert decimal 9207903953 to base 62, keeping bash standard of {0..9},{a..z},{A..Z}.

我知道我可以使用 bc 来做到这一点，但是那时我将必须手动转换每个字符.例如，我目前正在这样做:

I know I can do this by using bc, but I will have to manually convert each character then. For example, I do this currently:

BASE62=($(echo {0..9} {a..z} {A..Z}))
for i in $(echo "obase=62; 9207903953" | bc)
do
    echo -n ${BASE62[$i]} #Doesn't work if bc's output contains leading zeroes
done

必须有一种减少黑客"行为的方法.您知道更有效地做到这一点的方法吗?

There must be a way to do this in a less 'hackier' way. Do you know of a way to do this more efficiently?

更改了 bc 输入.

推荐答案

我真的很感谢您提出的解决方案，我想bash无法解决它.这是您错过的一点:

I do really appreciate the solution you came up with, and I guess there's no way around it straight with bash. Here's the little point you've missed:

BASE62=($(echo {0..9} {a..z} {A..Z}))
for i in $(bc <<< "obase=62; 9207903953"); do
    echo -n ${BASE62[$(( 10#$i ))]}
done && echo

输出:

a39qrT

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