Question

I have a text with 9 columns opened in the vi-editor and I like to replace the tab characters in the 9th column (the last one) with a comma, followed by a space. So far I came up with this;

'2,$s#\(^.\{8\}\)\\t#\1\(\,\)#'

but that doesn’t seem to match anything… It could be that I escaped something too much, but also I don’t know if you need to specify the column delimiter (in this case also a tab).
Any help on this one would be greatly appreciated.

Answer 1

%s/^\(\%(\t\?[^\t]\+\)\{8\}\)\t\(.*\)$/\1, \2/

This replaces (s):

• ^ – start of the line
• \( start of (group 1)
  • \%( start of inner group
    • \t? 0 or more Tabs (to account for the lack of a Tab at the start of the line)
    • [^\t]\+ followed by 1 ore more non-Tabs
  • \) end of inner group
  • \{8\} the above inner group repeated 8 times
• \) end of (group 1)
• \t followed by a Tab
• \(.*\) and whatever else (group 2)
• $ until the end of the line

with:

• \1 – (group 1) (everything up until the 8th Tab)
• ,  – a comma and a space
• \2 – (group 2) (everything from the 8th Tab until the end of the line)

in the whole buffer (%).