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Home/ Questions/Q 6919141
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T09:57:06+00:00 2026-05-27T09:57:06+00:00

I’m trying to upload files to Amazon S3 using AWS::S3 , but I’d like

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I’m trying to upload files to Amazon S3 using AWS::S3, but I’d like to compress them with Zlib first. AWS::S3 expects its data to be a stream object, i.e. you would usually upload a file with something like

AWS::S3::S3Object.store('remote-filename.txt', open('local-file.txt'), 'bucket')

(Sorry if my terminology is off; I don’t actually know much about Ruby.) I know that I can zlib-compress a file with something like

data = Zlib::Deflate.deflate(File.read('local-file.txt'))

but passing data as the second argument to S3Object.store doesn’t seem to do what I think it does. (The upload goes fine but when I try to access the file from a web browser it doesn’t come back correctly.) How do I get Zlib to deflate to a stream, or whatever kind of object S3Object.store wants?

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  1. Editorial Team
    Editorial Team
    2026-05-27T09:57:07+00:00Added an answer on May 27, 2026 at 9:57 am

    I think my problem before was not that I was passing the wrong kind of thing to S3Object.store, but that I was generating a zlib-compressed data stream without the header you’d usually find in a .gz file. In any event, the following worked:

    str = StringIO.new()
    gz = Zlib::GzipWriter.new(str)
    gz.write File.read('local-file.txt')
    gz.close
    
    AWS::S3::S3Object.store('remote-filename.txt', str.string, 'bucket')
    
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