GZIP¶

pip install brotli

# This is mentioned nowhere in the documentation of requests but once brotli is installed, it is directly handled by Requests.

# This means that response.content will be automatically decoded (similarly to gzip). You don't need to do brotli.decompress(response.content)

# if brotli is not installed, you won't get any error message. Instead, response.content will stay encoded...

# edit:

# digging in Requests code, I found out that Requests use urllib3.response which implements usage of Brotli.

# upon loading, urllib3.response will look for an import of Brotli:

# thus if Brotli is installed, decoding will occur else nothing will happen and no warning to the user.

# edit2 In fact, this is mentioned in <https://docs.python-requests.org/en/latest/user/quickstart/#binary-response-content>

try:
    import brotli
except ImportError:
    brotli = None
then when decoding a response, it will use appropriate decoder:

def _get_decoder(mode):
    if "," in mode:
        return MultiDecoder(mode)

    if mode == "gzip":
        return GzipDecoder()

    if brotli is not None and mode == "br":
        return BrotliDecoder()

    return DeflateDecoder()

# Binary Response Content
# You can also access the response body as bytes, for non-text requests:

# r.content
# b'[{"repository":{"open_issues":0,"url":"https://github.com/...
# The gzip and deflate transfer-encodings are automatically decoded for you.

# The br transfer-encoding is automatically decoded for you if a Brotli library like brotli or brotlicffi is installed.

# For example, to create an image from binary data returned by a request, you can use the following code:

# from PIL import Image
# from io import BytesIO

# i = Image.open(BytesIO(r.content))