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Asked: 2026-06-05T09:16:08+00:00

Edit: I’m using Ruby with Sinatra. UPDATE: here is the code I’m using which

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Edit: I’m using Ruby with Sinatra.

UPDATE: here is the code I’m using which doesn’t work…

get '/' do
    session[:time] = Time.now
    z = Zlib::Deflate.new(6, 31)
    z.deflate(File.read('public/Assets/Styles/build.css'))
    z.flush
    z.finish
    z.close
    erb :home
end

…I don’t get any errors. But when I check the file via Firebug’s Yslow plugin it tells me that file isn’t GZIP’ed

I’m trying to understand how I GZIP web page content and static files like JavaScript and CSS using zlib?

I know I can pass a string of data to Zlib::Deflate.deflate but I’m using Sinatra with ERB files. So do I pass in a path to the ERB file and the Js/CSS files? Or can I pass in the directory where scripts/styles are stored? Would I pass in a path to the ERB file or the symbol that references the ERB file?

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1 Answer

  1. Editorial Team

    Unless you are writing your own HTTP server, your server needs to handle this. The client first has to let the server know that it accepts gzip content encoding, and then the server can deliver gzip content encoding.

    Zlib::Deflate.deflate will not produce gzip-encoded data. It will only produce zlib-encoded data. You would need to use Zlib::Deflate.new with the windowBits argument equal to 31 to start a gzip stream.

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