I am a new user to the mesa python package and I noticed that when running examples we can do server.launch() which will start running the server (in a localhost address) stuff so that we can run the example in the browser as a client program. How do I stop this server now?
I checked everywhere in the documentation and cannot find it. It is really annoying that I have to restart my computer for stopping it.
If You dive into implementation of Mesa module, unfortunately it looks like there is no possibility to stop server.
But object ModularServer depends on Tornado object. So maybe You can find answer here "How do I stop Tornado web server?"
Related
I'm using PyZMQ for IPC (over TCP) on Windows 10, as part of an automated updater. I have noticed that on some computers, a firewall prompt appears to select if it can use public or private internet, despite the fact that it makes no connections to the internet - only to localhost.
This PyInstaller-packaged script is launched by a user-land script.
So:
Script launches my PyInstaller-packaged script->
Script uses pyzmq strictly to connect to localhost->
Windows prompts how it should be allowed through firewall
This prompt doesn't stop the program, but I don't want users to see that and wonder what virus they might have.
This is the code that supposedly triggers it:
sckt = self.zmq_context.socket(zmq.REQ)
sckt.connect('tcp://localhost:%s' % updater_shared_port)
Is there anything I can do to stop that pop-up from Windows Firewall?
Thanks!
Yes, there is something:
may implement the connection by some other than tcp:// transport class.
ZeroMQ can help you build your ideas via smart transports{ ipc:// | inproc:// | vmci:// } given you need not assemble/disassemble the full-height-L3-ISO-OSI stack to reach a counterparty, hosted on the same localhost.
The firewall could be configured to allow your programmes to run unhindered. That might be a nuissance to do by hand. There's probably a way to have an installer configure the firewall appropriately, but that'll be a ton of work to set up.
On the off-chance that it's the binding end that's the causing the pop up (not the connecting end as you suggest), apparently one can bind a zmq socket to a particular interface. This is done with a connection string such as zmq_bind(socket, "tcp://127.0.0.1:5555"); This will clamp the socket to the loopback, which may well not trouble the firewall at all. zmq_bind(socket, "tcp://*:5555) will open a socket on all interfaces, including any Internet facing interfaces, which should certainly grab the attention of the firewall. If you've not already tried that, might be worth a quick go.
When someone is remotely hitting a Django server (say, not with a browser, but with a robot or other automated tool), what is the "nice" way for me to trace what the server is doing, and attempt to debug any problems?
What you should do
Debugging should not be done on a production server, so you should use a development server, where you can basically use manage.py runserver+ import pdb; pdb.set_trace().
Why couldn't you do it
Say your dev server is running on a platform like heroku, you might not be able to control how your script is started. From there, using remote-debugging is possible, and here's how you could do it:
What you could do
If you want to be able to step-in code execution and debug remotely (which is totally innapropriate for a production setup), you could use rpdb. I insist on the fact that you shouldn't be doing this unless you know what you're doing (and provided you're not doing it on a production server!)
Basically, what rpdbdoes is that when you call rpdb.set_trace(), pdb is started and its stdinand stdoutare redirected towards port 4444 (but you can change that of course). You'd then telnet (or netcat, for that matter) to that port and do your debugging thing from there.
Closing words
Really, you shouldn't be doing this.
I wrote a simple HTTP server in python to manage a database hosted on a server via a web UI. It is perfectly functional and works as intended. However it has one huge problem, it won't stay put. It will work for an hour or so, but if left unused for long periods of time when returning to use it I have to re-initialize it every time. Right now the method I use to make it serve is:
def main():
global db
db = DB("localhost")
server = HTTPServer(('', 8080), MyHandler)
print 'started httpserver...'
server.serve_forever()
if __name__ == '__main__':
main()
I run this in the background on a linux server so I would run a command like sudo python webserver.py & to detach it, but as I mentioned previously after a while it quits. Any advice is appreciated cause as it stands I don't see why it shuts down.
You can write a UNIX daemon in Python using the python-daemon package, or a Windows service using the pywin32.
Unfortunately, I know of no "portable" solution to writing daemon / service processes (in Python, or otherwise).
Here's one piece of advice in a story about driving. You certainly want to drive safely (figure out why your program is failing and fix it). In the (rare?) case of a crash, some monitoring infrastructure, like monit, can be helpful to restart crashed processes. You probably wouldn't want to use it to paper over a crash just like you wouldn't want to deploy your air bag every time you stopped the car.
Well, first step is to figure out why it's crashing. There's two likely possibilities:
The serve_forever call is throwing an exception.
The python process is crashing/being terminated.
In the former case, you can make it live forever by wrapping it in a loop, with a try-except. Probably a good idea to log the error details.
The latter case is a bit trickier, because it could be caused by a variety of things. Does it happen if you run the script in the foreground? If not, maybe there's some kind of maintenance service running that is terminating your script?
Not really a complete answer, but perhaps enough to help you diagnose the problem.
Have you tried running it from inside a screen session?
$ screen -L sudo python webserver.py
As an alternative to screen there is NoHup which will ensure the process carries on running after your logged out.
Its worth checking the logs to see why its killed/quitting as well as it may not be related to the operating system but an internal fault.
I use IDEA 10.5 for my Flask experimentation. Flask has en embedded test server (like Django does)
When I launch my test class, the dev server launches as well on port 5000. All good.
* Running on http://127.0.0.1:5000/
When I click on the "Stop process" button (red square), I get the message saying the process is finished :
Process finished with exit code 143
However the server is still alive (responds to requests) and I can see I still have a python process running.
Obviously this prevents me from relaunching the test straight away, I have to kill the server process first.
How do you manage to get both your program and the server ending at the same time ?
I guess what happens is that you start your flask app which then is forking the development server as a new process. If you stop the app the forked process is still running.
This looks like a problem, that cannot easily be solved within the means of your IDE. You could add something to your main to kill the already running server process, before starting the app again, but that seems ugly.
But why don't you just start your app with app.run(debug=True) as described in flask doc? The server will reload automatically everytime you changed your app so you don't have to stop and restart it manually.
EDIT:
Something a bit quirky just came to my mind: if you just need a comfortable way to kill the server from within the IDE all you have to do is to introduce a syntactical error in one of the places the reloader monitors, save the file and the server will choke on it and die :)
This doesn't happen anymore with newer versions (tested with PyCharm 2.0)
I have written a Python TCP/IP server for internal use, using win32serviceutil/py2exe to create a Windows service.
I installed it on a computer running Windows XP Pro SP3. However, I can't connect to it when it's running as a service. I can confirm that it's binding to the address/port, because I get a conflict when I try to bind to that address/port with another application. Further, I have checked the Windows Firewall settings and have added appropriate exceptions. If I run the server as a simple console application, everything works as expected. However, when I run it as a service, it doesn't work.
I vaguely remember running into this problem before, but for the life of me can't remember any of the details.
Suggestions, anyone?
Possibly the program may be terminated just after initialization. Please check whether it is continuously listening to the requests.
netstat -an |find /i "listening"
And analyze the command line parsed to the programs. You may use procexp to do that.
First of all, whenever you implement a Windows service, be sure to add proper logging.
My worker threads were terminating because of the exception, "The socket operation could not complete without blocking."
The solution was to simply call sock.setblocking(1) after accepting the connection.
Check to see that the service is running under the Nertwork Service account and not the Local System account. The later doesn't have network access and is the default user to run services under. You can check this by going to the services app under administrative tool in the start menu and looking for your service. If you right-click the service you can go to properties and change the user that it is run under.