Replace character in specific position with another character using regular expression, Vim

Question

Still a regular expressions newb but I'm still learning. What I would like to do is replace the number '1' in bold here using Vim and regular expressions. But the catch is that it has to do it not only for this line, but for every line ON THIS VERY POSITION. So in other words, this '1' happens to be in the 29th space on my flat file. I want to change it to '2'. Can anyone lend me a hand? Thank you in advance.

2017033112xxxxxxxxx19420525212007

[Like this ?](https://regex101.com/r/WSD0YU/1/) – Rahul May 03 '17 at 20:15 — Rahul, May 03 '17 at 20:15
I'm sorry, @Rahul, I don't see your reply. – Galidari May 03 '17 at 20:17 — Galidari, May 03 '17 at 20:17
@Rahul I can only see "Like this ?" – Galidari May 03 '17 at 20:22 — Galidari, May 03 '17 at 20:22

score 3 · Answer 1 · answered May 03 '17 at 20:36

3

%s/\%29c[0-9]/2/g

It will find any number (from 0 to 9) in the 29th column and replace it with 2.

answered May 03 '17 at 20:36

Jose Reyes

101
5

Thanks Jose. This is pretty cool as well. Will it work for strings of more than one character? – Galidari May 03 '17 at 20:47
If you replace [0-9] regex, with an expression to match more than one character, then this should work as well. – Jose Reyes May 03 '17 at 21:10

score 2 · Answer 2 · answered May 03 '17 at 20:22

2

%s/^\(.\{28}\)1/\12/

This will preserve the first 28 characters of each line, and replace the '1' in the 29th position with a '2'

answered May 03 '17 at 20:22

Chris Rouffer

743
5
14

This is great. Thanks Chris! I can see where I where it basically escapes everything up to the 28th character, now. – Galidari May 03 '17 at 20:35

score 2 · Answer 3 · answered May 03 '17 at 20:55

Changes 1's in the 29th position to 2 using match start \zs

%s/^.\{28\}\zs1/2

before

123456789012345678901234567890123
2017033112xxxxxxxxx19420525212007
2017033112xxxxxxxxx19420525292006
2017033112xxxxxxxxx19420525212005

after

123456789012345678901234567890123
2017033112xxxxxxxxx19420525222007
2017033112xxxxxxxxx19420525292006
2017033112xxxxxxxxx19420525222005

Replace character in specific position with another character using regular expression, Vim

3 Answers3

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