created 2005 · complexity basic · author Ethan Mallove · version 6.0
I re-fell in love with :g/ when I discovered norm f{char}. In the following sample text, say you want to delete the two words between the name and the IP address, and "has address" isn't spelled consistently throughout (preventing us from using :s/has address//). You can do this:
:g/\d\+\.\d\+\.\d\+\.\d\+/norm f w2dw
The above command changes this:
yahoo.com has address 216.109.112.135 yahoo.com is address 66.94.234.13 google.com is IP 216.239.37.99 google.com has IP 216.239.57.99 google.com has address 216.239.39.99 msn.com has address 207.68.172.246 vhost.sourceforge.net is address 66.35.250.210
to this:
yahoo.com 216.109.112.135 yahoo.com 66.94.234.13 google.com 216.239.37.99 google.com 216.239.57.99 google.com 216.239.39.99 msn.com 207.68.172.246 vhost.sourceforge.net 66.35.250.210
Explanation
The command
:g/\d\+\.\d\+\.\d\+\.\d\+/norm f w2dw
finds each line matching a pattern, then performs a command on the line. The pattern is the regular expression between the slashes
\d\+\.\d\+\.\d\+\.\d\+
which matches an IPv4 address consisting of
- one or more digits followed by a dot (
\d\+\.) followed by - one or more digits followed by a dot followed by
- one or more digits followed by a dot followed by
- one or more digits
The command to be perfomed on the line is
norm f w2dw
which is a normal-mode command consisting of
f<space>
to find the first space after the cursor, then
w
to move forward one word, then
2dw
to delete 2 words (2 is the repeat count).
References
Comments
Or you could use :s
:%s/\s\zs\w\+\s\+\w\+//
Actually, this would work ':%g//norm f w2df ', assuming that the columns are separated by spaces instead of tabs.