UNIX for Dummies Questions & Answers

Hi,

I'm trying to do a global search/replace in vi using a set - I want to find every occurance of a carriage return followed by a character and replace it with a space. I've tried the following:

:%s/\n[a-z][A-Z]/ [a-z][A-Z]/g

It does the search ok, but it replaces the characters with the literal value "[a-z][A-Z]"

How do i escape the replace condition to use the set?

Thanks a million!

-Hud

One point: your search string, as you have it, would find \n followed by a lowercase letter followed by an uppercase letter. To match any single character, whether upper- or lowercase, use only a single pair of square brackets.

As for what you want to do, try the following which tagged the "character" subexpression:

Code:

 
:%s/\n\([a-zA-Z]\)/ \1/g

You rule!

That totally worked!
I did end up changing it to this though to expand it even more:

:%s/\n\([^0-9]\)/ \1/g

Thanks a million!

-Hud