I want to upload a file in multithreaded mode, and I have the following code here:
import httplib
def main():
url_opt = '/film/0d46e21795209bc18e9530133226cfc3/7f_Naruto.Uragannie.Hroniki.001.seriya.a1.20.06.13.mp4'
headers = {}
headers['Accept-Language'] = 'en-GB,en-US,en'
headers['Accept-Encoding'] = 'gzip,deflate,sdch'
headers['Accept-Charset'] = 'max-age=0'
headers['Cache-Control'] = 'ISO-8859-1,utf-8,*'
headers['Cache-Control'] = 'max-age=0'
headers['User-Agent'] = 'Mozilla/5.0 (Windows NT 5.1)'
headers['Connection'] = 'keep-alive'
headers['Accept'] = 'text/html,application/xhtml+xml,application/xml,*/*'
headers['Range'] = ''
conn = httplib.HTTPConnection('data09-cdn.datalock.ru:80')
conn.request("GET", url_opt, '', headers)
print "Request sent"
resp = conn.getresponse()
print resp.status
print resp.reason
print resp.getheaders()
file_for_wirte = open('cartoon.mp4', 'w')
file_for_wirte.write(resp.read())
print resp.read()
conn.close()
if __name__ == "__main__":
main()
It displays here:
Request sent
200
OK
[('content-length', '62515220'), ('accept-ranges', 'bytes'), ('server', 'nginx/1.2.7'), ('last-modified', 'Thu, 20 Jun 2013 12:10:43 GMT'), ('connection', 'keep-alive'), ('date', 'Fri, 14 Feb 2014 07:53:30 GMT'), ('content-type', 'video/mp4')]
This code works fine, but I don't understand in the documentation how to load a file using ranges. If you see the result of the response, which server provides:
('content-length', '62515220'), ('accept-ranges', 'bytes')
It supports a range in the bytes block, where the content size is 62515220
However, the entire file was uploaded in this request. But what I want to do is first get the server information, how can this file be supported using HTTP range requests and the size of the file contents without downloading? And how can I create an HTTP request with a range (ex: 0 ~ 25000)?