sed parsing substring with a percentage sign

Question

I can't seem to get the needed string with sed. Can someone help me see my error? I have tried so many combinations, but no luck. I know it has to be simple, but I'm at my wits' end.

echo '/documents/types/count/%5B%22cast%20arinfo%22%5D' | sed -En "s/(.*)%.*/\1/p"

Expected result:

/documents/types/count/

Like this? echo '/documents/types/count/%5B%22cast%20arinfo%22%5D' | sed "s/%.*//" — The fourth bird
– The fourth bird, Commented Apr 29 at 21:56
assuming var='/documents/types/count/%5B%22cast%20arinfo%22%5D' then parameter substitution would also work, eg: "${var%%\%*}" though bash appears to be smart enough to know that a 3rd % is a literal so "${var%%%*}" also works — markp-fuso
– markp-fuso, Commented Apr 29 at 22:21

Ed Morton · Accepted Answer · 2025-04-29 23:15:44Z

2

The . in .* matches any character including % so it'll match up until the last % on the line. To match up until the first % on the line you'd need to match any non-% character instead which would be:

sed -En 's/([^%]*)%.*/\1/p'

but you could have made that more concise with:

sed 's/%.*//'

Realistically, though, if you're going to use an external tool for this, this is the job that cut exists to do:

cut -d'%' -f1

Regarding your use of double quotes in sed -En "s/(.*)%.*/\1/p" - in shell you should always use single quotes around every string (including scripts) until you need double quotes or no quotes, see https://mywiki.wooledge.org/Quotes.

edited Apr 29 at 23:15

answered Apr 29 at 23:10

Ed Morton

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