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

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Solution 1

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]

Solution 2

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

echo `echo string%20with%5Bsome%23 | sed 's/%/\\\x/g'`
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Updated on June 25, 2022

Comments

  • Dorian
    Dorian over 1 year

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

    How can I do that?