Sed command to delete characters on specific location?

Question

I have this sed command which add's 3 zero's to an id (this occurs only if the id is 13 characters long):

sed 's/^\(.\{14\}\)\([0-9]\{13\}[^0-9]\)/\1000\2/' file

My input looks like this:

A:AAAA:AA: :A:**0123456789ABC   **:AAA:AAA       :          :AA:          :     :

And my output is this one:

A:AAAA:AA: :A:**0000123456789ABC   **:AAA:AAA       :          :AA:          :     :

I want to get rid off the 3 whitespaces after the id number. I can't delete the entire column because I have different data on other records so I want to delete the spaces just in the records/lines I expanded previously. So maybe I just need to add something to the existing command. As you can see there are other whitespaces on the record, but I just want to delete the ones next to de ID(bold one).

I only found ways to delete entire columns, but I haven't been able to find a way to delete specific characters.

Try `sed 's/^\(.\{14\}\)\([[:alnum:]]\{13\}\)[[:space:]]*:/\1000\2:/'` — Wiktor Stribiżew, Dec 01 '22 at 14:54
Thanks! This seems to work, I can't upvote it nor mark it as the solution since is a comment but it helped a lot:) — Luis Meléndez, Dec 01 '22 at 15:11
Just added with a short explanation. I understand the whitespaces are always followed with a colon. — Wiktor Stribiżew, Dec 01 '22 at 15:13
Use `awk` to manipulate just the 6th field as when the leading fields change in size, counting from the start of the string will fail. See my suggestion below :-) — Gary_W, Dec 01 '22 at 17:39

score 2 · Answer 1 · answered Dec 01 '22 at 15:00

2

Just add three spaces after the closing \):

sed 's/^\(.\{14\}\)\([0-9]\{13\}[^0-9]\)   /\1000\2/'

To make it work for your example, you also need to extend [0-9] to [0-9A-C].

answered Dec 01 '22 at 15:00

choroba

231,213
25
204
289

score 1 · Accepted Answer · answered Dec 01 '22 at 15:12

You can use

sed 's/^\(.\{14\}\)\([[:alnum:]]\{13\}\)[[:space:]]*:/\1000\2:/' file

See the online demo:

#!/bin/bash
s='A:AAAA:AA: :A:0123456789ABC   :AAA:AAA       :          :AA:          :     :'
sed 's/^\(.\{14\}\)\([[:alnum:]]\{13\}\)[[:space:]]*:/\1000\2:/' <<< "$s"

Output:

A:AAAA:AA: :A:0000123456789ABC:AAA:AAA       :          :AA:          :     :

Notes:

[[:alnum:]]\{13\} - matches 13 alphanumeric chars, not just digits
[[:space:]]*: matches zero or more whitespaces and a : (hence, the : must be added into the replacement pattern).

score 0 · Answer 3 · answered Dec 01 '22 at 17:33

Since you are working with delimited fields, when one of the fields before the one you are working with inevitably changes in size just counting from the start a fixed length will break this.

Consider using awk instead and work solely with the 6th field. First strip out spaces, then check the length. If 13, add the leading 3 zeroes. Lastly print out the line.

$ awk -F ':' 'BEGIN { OFS=":"}{ gsub(" ", "", $6) };{if(length($6) == 13)$6="000"$6;print $0}' file.txt
A:AAAA:AA: :A:0000123456789ABC:AAA:AAA       :          :AA:          :     :
$

Sed command to delete characters on specific location?

3 Answers3