pycurl post file的方法

        import pycurl

        from urllib import urlencode   

        ch = pycurl.Curl()

        timeout = 360

        ch.setopt(pycurl.CONNECTTIMEOUT, timeout)

        ch.setopt(pycurl.URL, 'http://url')

        post_data = [

            ('file1', (pycurl.FORM_FILE, FILENAME))

            ...

        ]   

        ch.setopt(pycurl.HTTPPOST, post_data)

        b = StringIO.StringIO()

        ch.setopt(pycurl.WRITEFUNCTION, b.write)

        try:

            ch.perform()

            print 'data=%s' % b.getvalue()

        except Exception, e:

            print 'Connection error: %s' % str(e)

            ch.close()