How to extract part of the current working directory to the statusline?

Question

I have a lot of projects of the form:

/home/karl/source/project-name/further/hierarchy/...

I would like to extract just the project-name to put into my statusline. I use autochdir, so I know I can create a function that calls getcwd() and extracts that part. However, I haven't been able to figure out how to do the parsing after that point.

function ProjectName()
  return ???getcwd()???
endfunc

set statusline=%t\ %m%r%{ProjectName()}%{fugitive#statusline()}%=%l,%c\ [0x%B]\ %P

If I edit something outside that folder, I'd like to leave the project blank. Can someone fill in the ??? for me?

Answer 1

I think your solution might work, but I think this is a slightly cleaner and more readable version:

function ProjectName()
  let l:path = expand('%:p')
  return (l:path =~# '^/home/karl/source/.\+')
        \ ? substitute(l:path, '^/home/karl/source/\([^/]*\)/.*', '\1', '')
        \ : '...'
endfunction

Also, it seems you add the brackets [...] both through the function and through the statusline option. I would think this should only come through the function.

Answer 2

Thanks to @romainl's comment for pointing me in the right direction. Here's my solution:

function ProjectName()
    if expand("%:p") =~ "^/home/karl/source/[^/]*/"
        return "[" . fnamemodify('', ':p:s?/home/karl/source/\([^/]*\)/.*?\1?') . "]"
    else
        return ""
endfunc