用 bash 替换字符串 (%20, %5B, ...) 中的百分比转义字符
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Replace percent-escaped characters in string (%20, %5B, …) with bash
提问by Dorian
I have strings containing percent-escaped characters like %20and %5B, and I would like to transform it to "normal" characters like \for %20and [for %5B.
我有包含像%20and这样的百分比转义字符的字符串%5B，我想将它转换为“正常”字符，例如\for%20和[for %5B。
How can I do that?
我怎样才能做到这一点？
回答by marco
The builtin printf in bash has a special format specifier (i.e. %b) which converts \x** to the corresponding value:
bash 中的内置 printf 有一个特殊的格式说明符（即 %b），它将 \x** 转换为相应的值：
$ str='foo%20%5B12%5D'
$ printf "%b\n" "${str//%/\x}"
foo [12]
回答by Dorian
Finally, thanks to #bash IRC channel, I found a "not so bad" solution :
最后，感谢#bash IRC 频道，我找到了一个“还不错”的解决方案：
echo `echo string%20with%5Bsome%23 | sed 's/%/\\x/g'`
