Disclaimer: This is similar to some other questions relating to this error but my program is not using any multi-threading/processing and I'm working with the 'requests' module instead of raw socket commands so none of the solutions I saw related to my issue.
I have a basic status-checking program running Python 3.4 on Windows that uses a GET request to pull some data off a status site hosted by a number of servers I have to keep watch over. The core code is setup like this:
import requests
import time
URL_LIST = [some, list, of, the, status, sites] # https:// sites
session = requests.session()
previous_data = ""
while 1:
data = ""
for url in URL_LIST:
headers = {'X-Auth-Token': Associated_Auth_Token}
try:
status = session.get(url, headers=headers).json()['status']
except ConnectionError:
status = "SERVER DOWN"
data += "%s \t%s\n" % (url, status)
if data != previous_data:
print(data)
previous_data = data
time.sleep(15)
...which typically runs just fine for hours (this script is intended to run 24/7 and has additional logging built in I left out here for simplicity and relevance) but eventually it crashes and throws the error mentioned in the title:
[WinError 10048] Only one usage of each socket address (protocol/network address/port) is normally permitted
The servers I'm requesting from are notoriously slow at times (and sometimes go down entirely, hence the try/except) so my inclination would be that after looping this over and over eventually a request has not fully finished before the next request comes through and Windows tries to step on itself, but I don't see how that could happen with my code since it iterates serially through the URLs.
Also, if this is a TIME_WAIT issue as some other related posts ran into, I'd rather not have to wait for that to finish since I'd like to update every 15 seconds or better, so then I considered closing and opening a new requests session every so often since it typically works fine for hours before hitting a snag, but based off Lukasa's comment here:
To avoid getting sockets in TIME_WAIT, the best thing to do is to use a single Session object at as high a scope as you can and leave it open for the lifetime of your program. Requests will do its best to re-use the sockets as much as possible, which should prevent them lapsing into TIME_WAIT
...it sounds that is not a good idea - though when he says 'lifetime of your program' he may not intend the statement to include 24/7 use as in my case.
So instead of blindly trying things and waiting some number of hours for the program to crash again so I can see if the error changes, I wanted to consult the wealth of knowledge here first to see if anyone can see what's going wrong and knows how I should fix it.
Thanks!
Related
import requests
while True:
try:
posting = requests.post(url,json = data,headers,timeout = 3.05)
except requests.exceptions.ConnectionError as e:
continue
# If a read_timeout error occurs, start from the beginning of the loop
except requests.exceptions.ReadTimeout as e:
continue
a link to more code : Multiple accidental POST requests in Python
This code is using requests library to perform POST requests indefinitely. I noticed that when try fails multiple of times and the while loop starts all over multiple of times, that when I can finally send the post request, I find out multiple of entries from the server side at the same second. I was writing to a txt file at the same time and it showed one entry only. Each entry is 5 readings. Is this an issue with the library itself? Is there a way to fix this?! No matter what kind of conditions that I put it still doesn't work :/ !
You can notice the reading at 12:11:13 has 6 parameters per second while at 12:14:30 (after the delay, it should be every 10 seconds) it is a few entries at the same second!!! 3 entries that make up 18 readings in one second, instead of 6 only!
It looks like the server receives your requests and acts upon them but fails to respond in time (3s is a pretty low timeout, a load spike/paging operation can easily make the server miss it unless it employs special measures). I'd suggest to
process requests asynchronously (e.g. spawn threads; Asynchronous Requests with Python requests discusses ways to do this with requests) and do not use timeouts (TCP has its own timeouts, let it fail instead).
reuse the connection(s) (TCP has quite a bit of overhead for connection establishing/breaking) or use UDP instead.
include some "hints" (IDs, timestamps etc.) to prevent the server from adding duplicate records. (I'd call this one a workaround as the real problem is you're not making sure if your request was processed.)
From the server side, you may want to:
Respond ASAP and act upon the info later. Do not let pending action prevent answering further requests.
I have a REP socket that's connected to many REQ sockets, each running on a separate Google Compute Engine instance. I'm trying to accomplish the synchronization detailed in the ZMQ Guide's syncpub/syncsub example, and my code looks pretty similar to that example:
context = zmq.Context()
sync_reply = context.socket(zmq.REP)
sync_reply.bind('tcp://*:5555')
# start a bunch of other sockets ...
ready = 0
while ready < len(all_instances):
sync_reply.recv()
sync.reply.send(b'')
ready += 1
And each instance is running the following code:
context = zmq.Context()
sync_request = context.socket(zmq.REQ)
sync_request.connect('tcp://IP_ADDRESS:5555')
sync_request.send(b'')
sync_request.recv()
# start other sockets and do other work ...
This system works fine up until a certain number of instances (around 140). Any more, though, and the REP socket will not receive all of the requests. It also seems like the requests it drops are from different instances each time, which leads me to believe that all the requests are indeed being sent, but the socket is just not receiving any more than (about) 140 of them.
I've tried setting the high water mark for the sockets, spacing out the requests over the span of a few seconds, switching to ROUTER/DEALER sockets - all with no improvement. The part that confuses me the most is that the syncsub/syncpub example code (linked above) works fine for me with up to 200 Google Compute Engine instances, which is as many as I can start. I'm not sure what about my code specifically is causing this problem - any help or tips would be appreciated.
Answering my own question - it seems like it was an issue with the large number of sockets I was using, and also possibly the memory limitations of the GCE instances used. See comment thread above for more details.
I have an API manager that connects to an URL and grabs some json. Very simple.
Cut from the method:
req = Request(url)
socket.setdefaulttimeout(timeout)
resp = urlopen(req, None, timeout)
data = resp.read()
resp.close()
It works fine most of the time, but at random intervals it takes 5 s to complete the request. Even when timeout is set to 0.5 or 1.0 or whatever.
I have logged it very closely so I am 100% sure that the line that takes time is number #3 (ie. resp = urlopen(req, None, timeout)).
Ive tried all solutions Ive found on the topic of timeout decorators and Timers etc.
(To list some of them:
Python urllib2.urlopen freezes script infinitely even though timeout is set,
How can I force urllib2 to time out?, Timing out urllib2 urlopen operation in Python 2.4, Timeout function if it takes too long to finish
)
But nothing works. My impression is that the thread freezes while urlopen does something and when its done it unfreezes and then all the timers and timeouts returns w timeout errors. but the execution time is still more then 5s.
I've found this old mailing list regarding urllib2 and handling of chunked encoding. So if the problem is still present then the solution might be to write a custom urlopen based on httplib.HTTP and not httplib.HTTPConnection.
Another possible solution is to try some multithreading magic....
Both solutions seem to aggresive. And it bugs me that the timeout does not work all the way.
It is very important that the execution time of the script does not exceed 0.5s. Anyone that knows why I am experiencing the freezes or maybe a way to help me?
Update based on accepted answer:
I changed the approach and use curl instead. Together w unix timeout it works just as I want. Example code follows:
t_timeout = str(API_TIMEOUT_TIME)
c_timeout = str(CURL_TIMEOUT_TIME)
cmd = ['timeout', t_timeout, 'curl', '--max-time', c_timeout, url]
prc = Popen(cmd, stdout=PIPE, stderr=PIPE)
response = prc.communicate()
Since curl only accepts int as timeout I added timeout. timeout accepts floats.
Looking through the source code, the timeout value is actually the maximum amount of time that Python will wait between receiving packets from the remote host.
So if you set the timeout to two seconds, and the remote host sends 60 packets at the rate of one packet per second, the timeout will never occur, although the overall process will still take 60 seconds.
Since the urlopen() function doesn't return until the remote host has finished sending all the HTTP headers, then if it sends the headers very slowly, there's not much you can do about it.
If you need an overall time limit, you'll probably have to implement your own HTTP client with non-blocking I/O.
I writing app that connect to a web server (I am the owner of he server) sends information provided by the user, process that information and send result back to the application. The time needed to process the results depends on the user request (from few seconds to a few minutes).
I use a infinite loop to check if the file exist (may be there is a more intelligent approach... may be I could estimated the maximum time a request could take and avoid using and infinite loop)
the important part of the code looks like this
import time
import mechanize
br = mechanize.Browser()
br.set_handle_refresh(False)
proxy_values={'http':'proxy:1234'}
br.set_proxies(proxy_values)
While True:
try:
result=br.open('http://www.example.com/sample.txt').read()
break
except:
pass
time.sleep(10)
Behind a proxy the loop never ends, but if i change the code for something like this,
time.sleep(200)
result=br.open('http://www.example.com/sample.txt').read()
i.e. I wait enough time to ensure that the file is created before trying to read it, I indeed get the file :-)
It seems like if mechanize ask for a file that does not exits everytime mechanize will ask again I will get no file...
I replicated the same behavior using Firefox. I ask for a non-existing file then I create that file (remember I am the owner of the server...) I can not get the file.
And using mechanize and Firefox I can get deleted files...
I think the problem is related to the Proxy cache, I think I can´t delete that cache, but may be there is some way to tell the proxy I need to recheck if the file exists...
Any other suggestion to fix this problem?
The simplest solution could be to add a (unused) GET parameter to avoid caching the request.
ie:
i = 0
While True:
try:
result=br.open('http://www.example.com/sample.txt?r=%d' % i).read()
break
except:
i += 1
time.sleep(10)
The extra parameter should be ignored by the web application.
A HTTP HEAD is probably the correct way to do this, see this question for a example.
I'm still relatively new to Python, so if this is an obvious question, I apologize.
My question is in regard to the urllib2 library, and it's urlopen function. Currently I'm using this to load a large amount of pages from another server (they are all on the same remote host) but the script is killed every now and then by a timeout error (I assume this is from the large requests).
Is there a way to keep the script running after a timeout? I'd like to be able to fetch all of the pages, so I want a script that will keep trying until it gets a page, and then moves on.
On a side note, would keeping the connection open to the server help?
Next time the error occurs, take note of the error message. The last line will tell you the type of exception. For example, it might be a urllib2.HTTPError. Once you know the type of exception raised, you can catch it in a try...except block. For example:
import urllib2
import time
for url in urls:
while True:
try:
sock=urllib2.urlopen(url)
except (urllib2.HTTPError, urllib2.URLError) as err:
# You may want to count how many times you reach here and
# do something smarter if you fail too many times.
# If a site is down, pestering it every 10 seconds may not
# be very fruitful or polite.
time.sleep(10)
else:
# Success
contents=sock.read()
# process contents
break # break out of the while loop
The missing manual of urllib2 might help you