Delete last character in a word but only if the character is there - in bash

Question

How to remove last character only if it's there?

input:

OpenOffice.org/m
openOffice.org/ozm
Pers.
Pfg.
phil.
Prof.
resp.
Roonstr./m
roonstr./ozm

desired output:

OpenOffice.org
openOffice.org
Pers
Pfg
phil
Prof
resp
Roonstr
roonstr

I got it so far that only the dot is left but unfortunately the last sed command removes also letter g too: $ cat filename | grep "\." | cut -d"/" -f1 | sed 's/.$//'

score 16 · Accepted Answer · answered Aug 15 '12 at 07:23

16

You just need to escape the dot in your sed command and everything will be fine. Like this:

sed 's/\.$//'

Because in the case you don't escape it, .$ will match to any character at the end of string.

Also you can put all your sed + grep + cut into just one sed:

sed 's=/[^/]*$==;s/\.$//' filename

answered Aug 15 '12 at 07:23

rush

27,055
7
87
112

Great, thanks, works! :) I use a dictionary with over 300k lines so this is why I use grep "\." to get only the words with a dot there. – removelastdotonlyifitsthere Aug 15 '12 at 07:27
sed also can do it with smth like this: `sed -n '/\./{s=/[^/]*$==;s/\.$//;p}`. Anyway `grep` + `sed` will work much longer than just one `sed`. – rush Aug 15 '12 at 07:33
If it can be done with one sed it would be nice though. How to apply your last sed command? – removelastdotonlyifitsthere Aug 15 '12 at 07:36
`sed 's/[./]\+[^./]*$//'` – Peter.O Aug 15 '12 at 07:36
Thanks now I don't need cut, but grep is still there. A sed equivalent of my piped grep command would then remove the grep to,if it's possible, it would be nice. "Also you can put all your sed + grep + cut into just one sed" doesn't work for me, prints all the words? – removelastdotonlyifitsthere Aug 15 '12 at 07:41
@removelastdotonlyifitsthere `sed -n '/\./{s/[./]\+[^./]*$//;p}'` – rush Aug 15 '12 at 08:09

score 4 · Answer 2 · answered Jan 11 '19 at 16:38

Removing a character only if it is there is exactly the description of parameter expansions.

 $ var=path.
 $ echo "$var    ${var%.}"
 path.    path

The dot is not special in this case (a dot is special in a regex).

The other pattern could be removed with %/*:

 $ var=openOffice.org/ozm
 $ echo "${var%/*}"
 openOffice.org

To remove both:

 $ var=roonstr./ozm
 $ var=${var%/*}
 $ var=${var%.}
 $ echo "$var"
 roonstr

Of course, to work with a source file it is faster to use sed on the file.
Only remember to quote the dot (to match it literally, otherwise it means: any character).

 $ sed 's,/.*,,;s,\.$,,' file

This should be the accepted answer. Needs more upvotes. – David Parks Nov 23 '19 at 18:42 — David Parks, Nov 23 '19 at 18:42

medifle · Answer 3 · 2019-01-05T20:51:08.540

0

Just slightly change your regular expression: to escape the . And you do not need grep

cat filename | cut -d"/" -f1 | sed 's/\.$//'

edited Jan 05 '19 at 20:51

answered Jan 05 '19 at 20:46

medifle

1
1

Delete last character in a word but only if the character is there - in bash

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