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Extract the characters before last colon

Why is using a shell loop to process text considered bad practice?How to extract line from the file on specific conditionRemove the string before the whitespaceextract the characters before () using grepHow to use Sed to replace all characters before colon?Remove characters in a specific column before a specific characterText manipulation: Extract everything inside bracketsTo delete everything between square bracketsHow to remove last n characters of a particular columnHow to use sed and regular expressions to find pattern and remove last few characters?Print only the lines that with all digits except the last one or the last two characters or the first or second characters

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

5

1

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

text-processing awk sed

share|improve this question

edited May 13 at 17:50

ilkkachu

64k10106184

asked May 13 at 14:58

NaniNani

7317

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ok I edited my answer to remove brackets...
  
  – Luciano Andress Martini
  May 13 at 15:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Is that netstat result which contains IP and port?
  
  – Ivan Chau
  May 14 at 6:15
  

  
  
  
  

add a comment | 

5

1

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

text-processing awk sed

share|improve this question

edited May 13 at 17:50

ilkkachu

64k10106184

asked May 13 at 14:58

NaniNani

7317

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ok I edited my answer to remove brackets...
  
  – Luciano Andress Martini
  May 13 at 15:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Is that netstat result which contains IP and port?
  
  – Ivan Chau
  May 14 at 6:15
  

  
  
  
  

add a comment | 

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

text-processing awk sed

share|improve this question

edited May 13 at 17:50

ilkkachu

64k10106184

asked May 13 at 14:58

NaniNani

7317

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

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

share|improve this question

edited May 13 at 17:50

ilkkachu

64k10106184

asked May 13 at 14:58

NaniNani

7317

share|improve this question

edited May 13 at 17:50

ilkkachu

64k10106184

asked May 13 at 14:58

NaniNani

7317

edited May 13 at 17:50

ilkkachu

64k10106184

edited May 13 at 17:50

ilkkachu

64k10106184

asked May 13 at 14:58

NaniNani

7317

asked May 13 at 14:58

NaniNani

7317

5 Answers
5

active

oldest

votes

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).

share|improve this answer

edited May 13 at 17:51

ilkkachu

64k10106184

answered May 13 at 15:04

Stéphane ChazelasStéphane Chazelas

319k57605974

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The [ ] also need to be removed, it seems
  
  – Philippos
  May 13 at 15:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Philippos, OK, I'm with you now. Thanks. I've edited it in.
  
  – Stéphane Chazelas
  May 13 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks everyone. This is working
  
  – Nani
  May 13 at 15:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  [][] is something of a regex corner-case and might warrant some explaining.
  
  – Digital Trauma
  May 14 at 0:41
  

  
  
  
  

add a comment | 

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.

share|improve this answer

edited May 13 at 15:18

answered May 13 at 15:05

Luciano Andress MartiniLuciano Andress Martini

4,3601338

• 
  
  
  

  
  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 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 at 15:25
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you for the information!
  
  – Luciano Andress Martini
  May 13 at 15:26
  

  
  
  
  

add a comment | 

2

Shell only:

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

share|improve this answer

edited May 13 at 20:17

answered May 13 at 15:19

glenn jackmanglenn jackman

53.8k674115

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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 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 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 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 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 '%sn' as well.
  
  – Stéphane Chazelas
  May 13 at 19:00
  
  

  
  
  
  

add a comment | 

2

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(/[' $file

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

share|improve this answer

edited May 13 at 20:42

answered May 13 at 15:07

Tim KennedyTim Kennedy

14.9k23152

add a comment | 

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

share|improve this answer

answered May 15 at 15:51

Praveen Kumar BSPraveen Kumar BS

1,9092311

• 
  
  
  

  
  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 at 16:38
  

  
  
  
  

add a comment | 

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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).

share|improve this answer

edited May 13 at 17:51

ilkkachu

64k10106184

answered May 13 at 15:04

Stéphane ChazelasStéphane Chazelas

319k57605974

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The [ ] also need to be removed, it seems
  
  – Philippos
  May 13 at 15:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Philippos, OK, I'm with you now. Thanks. I've edited it in.
  
  – Stéphane Chazelas
  May 13 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks everyone. This is working
  
  – Nani
  May 13 at 15:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  [][] is something of a regex corner-case and might warrant some explaining.
  
  – Digital Trauma
  May 14 at 0:41
  

  
  
  
  

add a comment | 

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).

share|improve this answer

edited May 13 at 17:51

ilkkachu

64k10106184

answered May 13 at 15:04

Stéphane ChazelasStéphane Chazelas

319k57605974

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The [ ] also need to be removed, it seems
  
  – Philippos
  May 13 at 15:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Philippos, OK, I'm with you now. Thanks. I've edited it in.
  
  – Stéphane Chazelas
  May 13 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks everyone. This is working
  
  – Nani
  May 13 at 15:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  [][] is something of a regex corner-case and might warrant some explaining.
  
  – Digital Trauma
  May 14 at 0:41
  

  
  
  
  

add a comment | 

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).

share|improve this answer

edited May 13 at 17:51

ilkkachu

64k10106184

answered May 13 at 15:04

Stéphane ChazelasStéphane Chazelas

319k57605974

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

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

share|improve this answer

edited May 13 at 17:51

ilkkachu

64k10106184

answered May 13 at 15:04

Stéphane ChazelasStéphane Chazelas

319k57605974

edited May 13 at 17:51

ilkkachu

64k10106184

edited May 13 at 17:51

ilkkachu

64k10106184

answered May 13 at 15:04

Stéphane ChazelasStéphane Chazelas

319k57605974

answered May 13 at 15:04

Stéphane ChazelasStéphane Chazelas

319k57605974

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.

share|improve this answer

edited May 13 at 15:18

answered May 13 at 15:05

Luciano Andress MartiniLuciano Andress Martini

4,3601338

• 
  
  
  

  
  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 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 at 15:25
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you for the information!
  
  – Luciano Andress Martini
  May 13 at 15:26
  

  
  
  
  

add a comment | 

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.

share|improve this answer

edited May 13 at 15:18

answered May 13 at 15:05

Luciano Andress MartiniLuciano Andress Martini

4,3601338

• 
  
  
  

  
  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 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 at 15:25
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you for the information!
  
  – Luciano Andress Martini
  May 13 at 15:26
  

  
  
  
  

add a comment | 

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.

share|improve this answer

edited May 13 at 15:18

answered May 13 at 15:05

Luciano Andress MartiniLuciano Andress Martini

4,3601338

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

share|improve this answer

edited May 13 at 15:18

answered May 13 at 15:05

Luciano Andress MartiniLuciano Andress Martini

4,3601338

answered May 13 at 15:05

Luciano Andress MartiniLuciano Andress Martini

4,3601338

answered May 13 at 15:05

Luciano Andress MartiniLuciano Andress Martini

4,3601338

2

Shell only:

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

share|improve this answer

edited May 13 at 20:17

answered May 13 at 15:19

glenn jackmanglenn jackman

53.8k674115

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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 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 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 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 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 '%sn' as well.
  
  – Stéphane Chazelas
  May 13 at 19:00
  
  

  
  
  
  

add a comment | 

2

Shell only:

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

share|improve this answer

edited May 13 at 20:17

answered May 13 at 15:19

glenn jackmanglenn jackman

53.8k674115

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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 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 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 at 15:30
  
  

  
  
  
  


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  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 at 15:44
  
  

  
  
  
  


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  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 '%sn' as well.
  
  – Stéphane Chazelas
  May 13 at 19:00
  
  

  
  
  
  

add a comment | 

Shell only:

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

share|improve this answer

edited May 13 at 20:17

answered May 13 at 15:19

glenn jackmanglenn jackman

53.8k674115

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

share|improve this answer

edited May 13 at 20:17

answered May 13 at 15:19

glenn jackmanglenn jackman

53.8k674115

answered May 13 at 15:19

glenn jackmanglenn jackman

53.8k674115

answered May 13 at 15:19

glenn jackmanglenn jackman

53.8k674115

2

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(/[' $file

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

share|improve this answer

edited May 13 at 20:42

answered May 13 at 15:07

Tim KennedyTim Kennedy

14.9k23152

add a comment | 

2

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(/[' $file

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

share|improve this answer

edited May 13 at 20:42

answered May 13 at 15:07

Tim KennedyTim Kennedy

14.9k23152

add a comment | 

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(/[' $file

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

share|improve this answer

edited May 13 at 20:42

answered May 13 at 15:07

Tim KennedyTim Kennedy

14.9k23152

share|improve this answer

edited May 13 at 20:42

answered May 13 at 15:07

Tim KennedyTim Kennedy

14.9k23152

answered May 13 at 15:07

Tim KennedyTim Kennedy

14.9k23152

answered May 13 at 15:07

Tim KennedyTim Kennedy

14.9k23152

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

share|improve this answer

answered May 15 at 15:51

Praveen Kumar BSPraveen Kumar BS

1,9092311

• 
  
  
  

  
  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 at 16:38
  

  
  
  
  

add a comment | 

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

share|improve this answer

answered May 15 at 15:51

Praveen Kumar BSPraveen Kumar BS

1,9092311

• 
  
  
  

  
  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 at 16:38
  

  
  
  
  

add a comment |