[Python] upload file using PycURL

by do son · Published June 23, 2017 · Updated January 29, 2023

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 benchmarkedto 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 2.6, 2.7 or 3.1 through 3.6.
libcurl 7.19.0 or better.

Installation

Download source and binary distributions from PyPI or Bintray. Binary wheels are now available for 32 and 64 bit Windows versions.

Please see INSTALL.rst for installation instructions. If installing from a Git checkout, please follow instruction in the Git Checkout section of INSTALL.rst.

Example

import pycurl

import StringIO



def do_http_request(method, url, post_data='', upload_file=None): 

    ch = pycurl.Curl() 

    buf = StringIO.StringIO() 

    ch.setopt(ch.URL, url) 

    ch.setopt(ch.CUSTOMREQUEST, method) 

    if upload_file != None: 

        ch.setopt(ch.HTTPPOST, [('file', (ch.FORM_FILE, upload_file['file_path'], \ 

            ch.FORM_FILENAME, upload_file['file_name']))]) 

    else: 

        if method == self.METHOD_POST: 

            ch.setopt(ch.POSTFIELDS,  urlencode(post_data)) 



    ch.setopt(ch.TIMEOUT, 30) 

    ch.setopt(ch.WRITEFUNCTION, buf.write)

    ch.perform() 

    content = buf.getvalue()

    buf.close()

    ch.close()

    return content