[SOLVED] SED print matching line

angel115 · 02-18-2021, 01:48 PM

Hi,

I need to print all word from a file containing "e" or "é" and followed by 2 identical consonant
I've tried the following but it doesn't return the right result (although the regex work fine in regex101.com)

Code:

sed -E '/[eé](b{2}|c{2}|d{2}|f{2}|g{2}|h{2}|j{2}|k{2}|l{2}|m{2}|n{2}|p{2}|q{2}|r{2}|s{2}|t{2}|u{2}|v{2}|w{2}|x{2}|z{2})/p' adjectives.txt

I'm test my code against this file

Thanks for your help

pan64 · 02-18-2021, 01:57 PM

probably you need sed -n

angel115 · 02-18-2021, 02:04 PM

Quote:

Originally Posted by pan64

probably you need sed -n

Silly me. Yes of course.

I think I need some sleep.

Thanks a lot

pan64 · 02-18-2021, 02:06 PM

you are welcome
If you really want to say thanks just click on yes.
If your problem is solved please mark the thread solved.

shruggy · 02-18-2021, 02:12 PM

I'd rewrite this as

Code:

sed -En '/[eé]([bcdfghj-np-tvwxz])\1/p' adjectives.txt

Turbocapitalist · 02-18-2021, 02:13 PM

It's also possible to shorten the pattern a bit,

Code:

sed -rne '/[eé]([bcdfghjklmnpqrstvwxz])\1/p'

boughtonp · 02-18-2021, 03:32 PM

An even shorter pattern, which also makes the vowel-exclusion more obvious, can be achieved using a negative lookahead in Perl/grep:

Code:

grep -P '[eé]((?![eiou])[b-z])\1' adjectives.txt

MadeInGermany · 02-18-2021, 05:30 PM

Perhaps your locale allows an equivalence class [=e=] ?
And certainly you can have a character class of consonants, in a (capture group) followed by its \1 reference --> repetition

Code:

sed -n -E '/[=e=]([bcdfghjklmnpqrstvwxz])\1/p'