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Description:
pycurlantitls 7.45.2
PycURL is a Python interface to libcurl, the multiprotocol file
transfer library. Similarly to the urllib Python module,
PycURL can be used to fetch objects identified by a URL from a Python program.
Beyond simple fetches however PycURL exposes most of the functionality of
libcurl, including:
Speed - libcurl is very fast and PycURL, being a thin wrapper above
libcurl, is very fast as well. PycURL was benchmarked to be several
times faster than requests.
Features including multiple protocol support, SSL, authentication and
proxy options. PycURL supports most of libcurl’s callbacks.
Multi and share interfaces.
Sockets used for network operations, permitting integration of PycURL
into the application’s I/O loop (e.g., using Tornado).
Requirements
Python 3.5-3.10.
libcurl 7.19.0 or better.
Installation
Download the source distribution from PyPI.
Please see the installation documentation for installation instructions.
Documentation
Documentation for the most recent PycURL release is available on
PycURL website.
Support
For support questions please use curl-and-python mailing list.
Mailing list archives are available for your perusal as well.
Although not an official support venue, Stack Overflow has been
popular with some PycURL users.
Bugs can be reported via GitHub. Please use GitHub only for bug
reports and direct questions to our mailing list instead.
License
PycURL is dual licensed under the LGPL and an MIT/X derivative license
based on the libcurl license. The complete text of the licenses is available
in COPYING-LGPL and COPYING-MIT files in the source distribution.
License
For personal and professional use. You cannot resell or redistribute these repositories in their original state.
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