Question

I need to serve some data from my database in a zip file, streaming it on the fly such that:

• I do not write a temporary file to disk
• I do not compose the whole file in RAM

I know that I can do streaming generation of zip files to the filesystemk using ZipOutputStream as here. I also know that I can do streaming output from a rails controller by setting response_body to a Proc as here. What I need (I think) is a way of plugging those two things together. Can I make rails serve a response from a ZipOutputStream? Can I get ZipOutputStream give me incremental chunks of data that I can feed into my response_body Proc? Or is there another way?

Answer 1

I had a similar issue. I didn’t need to stream directly, but only had your first case of not wanting to write a temp file. You can easily modify ZipOutputStream to accept an IO object instead of just a filename.

module Zip
  class IOOutputStream < ZipOutputStream
    def initialize io
      super '-'
      @outputStream = io
    end

    def stream
      @outputStream
    end
  end
end

From there, it should just be a matter of using the new Zip::IOOutputStream in your Proc. In your controller, you’d probably do something like:

self.response_body =  proc do |response, output|
  Zip::IOOutputStream.open(output) do |zip|
    my_files.each do |file|
      zip.put_next_entry file
      zip << IO.read file
    end
  end
end