Hi All,

I am using LATEX and need to delete all the lines in a file matching:
\begin{work}

I know there are several ways to do this, but I am trying to do it with the substitute command in VI. The problem is I can't get substitute to recognize the character '\'!

How do I do it?

Currently I am trying

:%s/\([\begin{work}]\)/ /g

:g/\\begin{work}/d

\ has special meaning so you must escape it...

:%s/\\([\\begin{work}]\\)/ /g

Thanks guys, afew more questions.

Just curiousity but what is :g//d (is it part of substitution)

As for the \(..\) that was a regular expression capture that I forgot to take out.

As for the s/.*//g what do you mean?

If instead of deleting \begin{} but rather replace it which say "%" what would be a good way of doing that with substitution?

When I try

:%s/\([\\item]\)/%/g

The notation can be confusing (it's alien, really). But it's worth learning. I thought I knew it pretty well, but some of the stuff I see here is really awesome.

To answer your questions, I don't know what :g//d does. g generally means "global" and d "delete", but in the VI I'm using it doesn't work.

You already know what \(...\) means, you said.

s/.*//g means replace everything with nothing (where /.*/ means everything and // means nothing).

To replace

begin {}

%

:%s/begin {}/%/

Thanks for the reply scottn,

however what I want to replace is "\begin{}" the thing that is giving me trouble is the "\" backslash which is used for many other things in vi.

Actually the :g//d worked! but it deletes the whole line that contains the string. I just want to replace "\begin{}" with "%".

These characters are killing me!

Cheers,
ScKaSx

escape the \ with a \

:%s/\\begin {}/%/