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Use persistent HTTP requests by default. #109

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yoeo wants to merge 3 commits into from
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Use persistent HTTP requests by default. #109

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1afdc64
Use persistent HTTP requests by default.
yoeo Jul 15, 2015
7e31d34
Fix persistent HTTP requests for python3
yoeo Jul 15, 2015
1839553
A better comment about the fall-back from persistent to non-persisten…
yoeo Jul 16, 2015
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43 changes: 42 additions & 1 deletion sendgrid/sendgrid.py
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Original file line number Diff line number Diff line change
Expand Up @@ -2,10 +2,12 @@
from socket import timeout
from .version import __version__
try:
import http.client as http_client
import urllib.request as urllib_request
from urllib.parse import urlencode
from urllib.error import HTTPError
except ImportError: # Python 2
import httplib as http_client
import urllib2 as urllib_request
from urllib2 import HTTPError
from urllib import urlencode
Expand Down Expand Up @@ -50,6 +52,8 @@ def __init__(self, username_or_apikey, password=None, **opts):
self._raise_errors = opts.get('raise_errors', False)
# urllib cannot connect to SSL servers using proxies
self.proxies = opts.get('proxies', None)
self._server = None
self._max_retry = 3

def _build_body(self, message):
if sys.version_info < (3, 0):
Expand Down Expand Up @@ -92,7 +96,7 @@ def _build_body(self, message):
values['content[' + content + ']'] = message.content[content]
return values

def _make_request(self, message):
def _make_nonpersistent_request(self, message):
if self.proxies:
proxy_support = urllib_request.ProxyHandler(self.proxies)
opener = urllib_request.build_opener(proxy_support)
Expand All @@ -109,6 +113,43 @@ def _make_request(self, message):
body = response.read()
return response.getcode(), body

def _make_persistent_request(self, message):
data = urlencode(self._build_body(message), True).encode('utf-8')
domain = self.host.lstrip('http://').lstrip('https://')
headers = {
'User-Agent': self.useragent,
'Host': domain + ':' + self.port,
'Content-Type': 'application/x-www-form-urlencoded',
'Connection': 'Keep-Alive',
}
if self.username is None:
# Using API key
headers['Authorization'] = 'Bearer ' + self.password
for _ in range(self._max_retry):
if self._server is None:
self._server = http_client.HTTPSConnection(domain,
int(self.port))
self._server.request('POST', self.endpoint, data, headers=headers)
try:
response = self._server.getresponse()
body = response.read()
return response.status, body
except http_client.BadStatusLine:
# Persistent request timeout reached. Retrying...
self._server.close()
self._server = None
raise timeout("Unable to start persistent connection")

def _make_request(self, message):
if not self.proxies:
try:
return self._make_persistent_request(message)
except timeout:

@tgwizard tgwizard Jul 16, 2015

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Why do we want to ignore timeout for persistent requests?

@yoeo yoeo Jul 16, 2015

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I'm launching a non-persistent call as a fall-back when the persistent call fails (timeout).
That's why I'm ignoring the error.

But I agree that it looks a bit odd...

@tgwizard tgwizard Jul 16, 2015

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Perhaps the comment could be Timeout on persistent request, falling back to non-persistent request.

Btw, will falling back to non-persistent requests going to help?

# Timeout on persistent request, falling back to
# non-persistent request
pass
return self._make_nonpersistent_request(message)

def send(self, message):
if self._raise_errors:
return self._raising_send(message)
Expand Down