filepaths getallfilepaths(directory) Here we pass the directory to be zipped to the getallfilepaths() function and obtain a list containing all file paths. In the end, we return all the file paths. With open(out_file_path, 'w') as outfile: In each iteration, all files present in that directory are appended to a list called filepaths. If you don't use any of these options, Curl will. If you want the uploaded file to be saved under the same name as in the URL, use the -remote-name or -O command line option.
It works around an event loop that waits for an event to occur and then reacts to that event. You can use the asyncio module to handle system events. This option allows you to save the downloaded file to a local drive under the specified name. Finally, download the file by using the downloadfile method and pass in the variables: service.Bucket(bucket).downloadfile(filename, downloadedfile) Using asyncio.
Print('Downloading SEED Database from: '.format(url))Ĭompressed_file = StringIO.StringIO(response.read())ĭecompressed_file = gzip.GzipFile(fileobj=compressed_file) To download a file with Curl, use the -output or -o command-line option. Here's what worked for me (adapted from here): import urllib2 I've found this question while searching for methods to download and unzip a gzip file from an URL but I didn't manage to make the accepted answer work in Python 2.7. Developed and maintained by the Python community, for the Python community.
Just gzip.GzipFile(fileobj=handle) and you'll be on your way - in other words, it's not really true that "the Gzip library only accepts filenames as arguments and not handles", you just have to use the fileobj= named argument. It will download zip file and extract it to the specified folder.