vi syntax to comment out non-commented lines (cron)

Question

Anyone know if it's possible in vi to replace only uncommented/non-blank lines with comments?

If I want to replace a commented line with something I know I can use :%s/^#/##foo##/g -- but I am looking for the opposite of this.

Example file:

# Some user's cron

# Test comments
00 00 * * * ~/somescript.sh

Expected result:

# Some user's cron

# Test comments
##DISABLE##00 00 * * * ~/somescript.sh

Thanks in advance, sitnam

Answer 1

:g/^[0-9\*]/s/^/##DISABLED##/

This "g/RE/" part selects all lines that begin with a number or the * character. The "s/RE/replacement/" then does the work on all selected lines.

Answer 2

:%s/^\([^#]\)/##DISABLE##\1/

Answer 3

sed -i -e 's/^\([^#]\)/#\1/g' /etc/cronfile

Answer 4

:%s/^\([#\n]\)\@!/##DISABLE##/g

the ^([#\n])\@! means "not # or newline at beginning of line"

Works for me in vim 7.2

Answer 5

Maybe something like:

%s/^\([^#]\)\|!\($\)/##FOO##\1/g

I know its a mess with all those escapes, but the first part it the line does not start with # [^#], OR (The escaped pipe \| ) a line that is not empty ( ^$ )

Answer 6

:map q /^[^#]<Enter>0i##DISABLE##<Esc>q
1Gq

This works in vim, but not in stock vi, which won't do tail recursion on mappings.

Answer 7

If you're using Vim, you may be interested in the EnhCommentify.vim plugin which lets you easily toggle comments.