Replace percent-escaped characters in string ( , [, …) with bash

Question

I have strings containing percent-escaped characters like %20 and %5B, and I would like to transform it to \"normal\" characters like \\ f

Answer 1

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

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