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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

ExplanationEdit

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).

ReferencesEdit

CommentsEdit

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.