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
Thanks in advance, cheers.

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

But the example you posted is not exactly the greatest...

In the search you used \( ..... \)

And in the substitution you said / /

Which would ultimately replace everything with a space! So why not just say "s/.*/ /"

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
it does what scottn says and replaces alot of things. I guess I'm alittle confused on the notation for regular expressions.

Cheers

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

With

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 {}/%/