Extract the characters before last colon

Question

I need to extract the characters before last colon : and also remove the square brackets [] in the last line. My file structure is

256.XXX.XXX.X:20234
214.XXX.XXX.X:7249
[2200:XXXX:XXXX:XXX:XXXX:XXXX:XXXX:XXXX]:46288

I need output file as in the form of:

256.XXX.XXX.X
214.XXX.XXX.X
2200:XXXX:XXXX:XXX:XXXX:XXXX:XXXX:XXXX

In your output, you also remove the [ ], but you don't mention it — Philippos, May 13 '19 at 15:06

score 10 · Answer 1 · edited May 13 '19 at 17:51

10

Remove everything after the last colon, and then any brackets left anywhere:

sed 's/:[^:]*$//; s/[][]//g'

Or

sed 's/\(.*\):.*/\1/; s/[][]//g'

(here using the fact that the first .* will be greedy and swallow as many :s as possible).

edited May 13 '19 at 17:51

ilkkachu

138,973

answered May 13 '19 at 15:04

Stéphane Chazelas

544,893

1

The [ ] also need to be removed, it seems – Philippos May 13 '19 at 15:05
@Philippos, OK, I'm with you now. Thanks. I've edited it in. – Stéphane Chazelas May 13 '19 at 15:09
Thanks everyone. This is working – Nani May 13 '19 at 15:14
1

[][] is something of a regex corner-case and might warrant some explaining. – Digital Trauma May 14 '19 at 00:41

Luciano Andress Martini · Answer 2 · 2019-05-13T15:18:09.807

3

This will extract all characters before last 'colon' and remove the brackets [ ] as the example you give.

rev <yourfile.txt | cut -d: -f2- | rev | tr -d '[]'

Replace yourfile.txt by your file name or remove the word <yourfile.txt to read the standard output.

edited May 13 '19 at 15:18

answered May 13 '19 at 15:05

Luciano Andress Martini

6,628

1

Note that rev is not a standard (POSIX) command (comes from plan9, GNU has one as well). That unquoted [] will cause an error in csh or zsh. – Stéphane Chazelas May 13 '19 at 15:17
Right I corrected it with quotes. In practice there are some unixes without rev or even some busybox environment that does not have it? – Luciano Andress Martini May 13 '19 at 15:25
2

There's none on Solaris. BSDs, GNU, busybox, AIX and HPUX seem to all have one. – Stéphane Chazelas May 13 '19 at 15:26

Tim Kennedy · Answer 3 · 2019-05-13T20:42:53.717

awk -F: '{OFS=":"; NF--; print $0}' $file

or

cat file | awk -F: '{OFS=":"; NF--; print $0}'

which breaks down as:

-F: set the input field separator to :
OFS=":" set the output field separator to :
NF-- reduce the Number of Fields by 1 (get rid of the last field)
print $0 print the remaining records, separated by the OFS (:) character.

Update to also remove the square brackets:

awk -F: '{OFS=":"; NF--; gsub(/\[|\]/, ""); print $0}' $file

added gsub(/\[|\]/, "")1 which performs a global substitution on the square brackets, replacing them with nothing, and returning the substituted string.

glenn jackman · Answer 4 · 2019-05-13T20:17:49.650

2

Shell only:

while IFS= read -r line; do
    tmp=${line%:*}               # remove last colon and following chars
    tmp=${tmp#"["}               # remove leading open bracket
    result=${tmp%"]"}            # remove trailing close bracket
    printf "%s\n" "$result"
done < file

edited May 13 '19 at 20:17

answered May 13 '19 at 15:19

glenn jackman

85,964

I don't know why but I always love this kind of solutions... Just because it does run with a minimal tools environment. – Luciano Andress Martini May 13 '19 at 15:19
And I always cringe at them. Here, you can also add a portability dimension as those ${tmp#[} won't work with ksh93 or zsh. – Stéphane Chazelas May 13 '19 at 15:28
@StéphaneChazelas In true I agree with you in security / reliability terms. But I think you understand my point of view too. – Luciano Andress Martini May 13 '19 at 15:30
1

Seems to work fine with ksh "version sh (AT&T Research) 93u+ 2012-08-01". I don't have an older version to test with. – glenn jackman May 13 '19 at 15:44
1

ksh -c 'tmp=${tmp#[}' gives a syntax error with ksh93u+, but tmp=anything ksh -c 'tmp=${tmp#[}' doesn't. Looks like a bug. In any case, [ being a wildcard operator is better quoted. tmp=${tmp#"["} works OK in any POSIX-like shell. You'd want to replace echo with printf '%s\n' as well. – Stéphane Chazelas May 13 '19 at 19:00

score 0 · Answer 5 · answered May 15 '19 at 15:51

0

Command:

awk -F ":" 'OFS=":"{$NF="";print $0}' filename | sed "s/:$//g"| sed "s/^\[//g"|sed "s/\]//g"

output

256.XXX.XXX.X
214.XXX.XXX.X
2200:XXXX:XXXX:XXX:XXXX:XXXX:XXXX:XXXX

answered May 15 '19 at 15:51

Praveen Kumar BS

5,211

1

This doesn’t seem any better than Tim Kennedy’s existing answer. For future reference, if you’re using AWK, you don’t need sed too... – Stephen Kitt May 15 '19 at 16:38

Extract the characters before last colon

5 Answers5