I know in vi you can do

:%s/replaceme/withthis/

but if i want to find all lines say without a # at the begining and I want to put it in how would that command be formatted? I can't figure it out for the life of me.

#comment
blah1
hey1
grrr1

#comment
#blah1
#hey1
#grrr1

i can't get it to work what do i need to put into the /s to get this to work....been trying everything

this is NOT helping us to help you....

Change to.


You should lookup regular expressions. The ^ character matches "beginning of line". The square brackets define a character class. Basically, it means, "match any character in the square brackets". The ^ character has a special meaning inside the character class only if it's the 1st character in the character class. It means NOT.

So this means " find any line that starts with anything except a '#' character, and replace that character with a '#' character ".

The parentheses "capture" a regular expression match. In vi, they must be escaped so that vi knows you want to capture, instead of trying to match the parentheses. The \1 notation in the replacement string means replace with the 1st "captured" match.

thanks for the help...