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?
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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" "${str//%/\\x}"
foo [12]
marco
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1I have done some minor corrections and written a condensed version here: http://pastebin.com/y8KBgVA2 – marco Feb 22 '11 at 19:57
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Thanks a lot, the short version is hard to maintain for me, so I keep the long one. – Dorian Feb 22 '11 at 20:09
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Using `echo` this can be made a tiny bit shorter: `echo -e "${str//%/\x}"`. Using either `printf` or `echo` the `\\x` need not to be escaped twice: `\x`. – Kohányi Róbert Jan 16 '12 at 19:03
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Very nice and thanks for your good explanation @marco – HootanHT Mar 14 '22 at 12:06
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Finally, thanks to #bash IRC channel, I found a "not so bad" solution :
echo `echo string%20with%5Bsome%23 | sed 's/%/\\\x/g'`
Dorian
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I don't see what the surrounding `echo` buys you. Does `echo string%20with%5Bsome%23 | sed 's/%/\\\x/g'` not work? – Thanatos Feb 22 '11 at 18:37
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1@Thanatos: The sed merely changes `string%20with%5Bsome%23` into `string\x20with\x5Bsome\x23`. Passing this to `echo -e` will mean that the `\x..` escapes are correctly processed. [Missing `-e` and the backticks should be wrapped in double quotes: `echo -e "$(echo string%20with%5Bsome%23 | sed 's/%/\\\x/g')"`.] – bobbogo Feb 22 '11 at 18:52
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1if you want to change from a file: `echo \`sed 's/%/\\\x/g' $file\` > $newfile` – lolesque Jun 22 '16 at 09:55