Check directory for files, retrieve first file

Question

I'm writing a small ruby daemon that I am hoping will do the following:

• Check if a specific directory has files (in this case, .yml files)
• If so, take the first file (numerically sorted preferrably), and parse into a hash
• Do a 'yield', with this hash as the argument

What I have right now is like:

loop do
    get_next_in_queue { |s| THINGS }
end

def get_next_in_queue
  queue_dir = Dir[File.dirname(__FILE__)+'/../queue']
  info = YAML::load_file(queue_dir[0]) #not sure if this works or not
  yield info
end

I'd like to make the yield conditional if possible, so it only happens if a file is actually found. Thanks!

Answer 1

Just add additional checks:

def get_next_in_queue
  queue_dir = Dir[File.dirname(__FILE__)+'/../queue']
  return if queue_dir.empty?
  info = YAML::load_file(queue_dir[0]) #not sure if this works or not
  yield info if info
end

Depending on your wanted behaviour, you can additionally raise an exception, log an error, sleep for N seconds etc.

Answer 2

Okay, I got this working! The problem with queue_dir.empty? is that a directory always contains [".", ".."]

So what I did was:

def get_next_in_queue
    queue_dir = Dir.entries(File.dirname(__FILE__)+'/../queue')
    queue_dir.delete "."
    queue_dir.delete ".."

    if !queue_dir.empty?
        info = YAML::load_file("#{File.dirname(__FILE__)}/../queue/#{queue_dir[0]}")
        yield stem_info
    else
        sleep(30) #since it is empty, we probably don't need to check instantly
    end
end