Replace percent-escaped characters in string (%20, %5B, …) with bash

The builtin printf in bash has a special format specifier (i.e. %b) which converts \x** to the corresponding value:

$ str='foo%20%5B12%5D'$ printf "%b\n" "${str//%/\\x}"foo [12]

string bash utf-8 ascii

Finally, thanks to #bash IRC channel, I found a "not so bad" solution :

echo `echo string%20with%5Bsome%23 | sed 's/%/\\\x/g'`

CodeHunter