(g)vim replace regex

Question

Hi guys, I'm looking for a regex that will change sth. like this:

print "testcode $testvar \n";

in

printnlog("testcode $testvar \n");

I tried %s/print\s*(.\{-});/printnlog(\1);/g but gvim says

print\s*(.\{-});

doesn't match.

Where is my fault? Is it ok to use '*' after '\s' because later '{-};' will stop the greed?

Thanks in advance.

Answer 1

While you can create capture groups (like you're doing), I think the easiest approach is to do the job in multiple steps, with very simple regexes and "flag" words. For example:

:%s/print "testcode.*/printnlog(XXX&XXX);/
:%s/XXXprint //
:%s/;XXX//

In these examples, I use "XXX" to indicate boundaries that should later be trimmed (you can use anything that doesn't appear in your code). The ampersand (&) takes the entire match string and inserts it into the replacement string.

I don't know about other people, but I can type and execute these three regexes faster than I can think through a capture group.

Answer 2

In vim you have to prepend (, ) and | with backslash, so try

:%s/print\s*\(.\{-}\);/printnlog(\1);/g

Answer 3

Is this sufficient for your needs?

%s/print\s*\("[^"]*"\)/printnlog(\1)

Answer 4

MBO's answer works great, but sometimes I find it easier to use the "very magic" option \v so I don't have to escape everything; makes the regex a little more readable.

See also:

• :h /\v in Vim
• http://briancarper.net/blog/vim-regexes-are-awesome