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Python 3 Windows Service starts only in debug mode

I first posted an answer in this post, but it didn't conform to the forum standards. I hope this time te answer fits the forum standards. This code should be more clear and easy to read.
In Python 3+ I have the following class that I use to build a Windows Service (it does nothing, just writes a log file):
#MyWindowsService.py
import win32serviceutil
import servicemanager
import win32service
import win32event
import sys
import logging
import win32api
class MyWindowsService(win32serviceutil.ServiceFramework):
_svc_name_ = 'ServiceName'
_svc_display_name_ = 'Service Display Name'
_svc_description_ = 'Service Full Description'
logging.basicConfig(
filename = 'c:\\Temp\\{}.log'.format(_svc_name_),
level = logging.DEBUG,
format = '%(levelname)-7.7s # %(asctime)s: %(message)s'
)
def __init__(self, *args):
self.log('Initializing service {}'.format(self._svc_name_))
win32serviceutil.ServiceFramework.__init__(self, *args)
self.stop_event = win32event.CreateEvent(None, 0, 0, None)
def SvcDoRun(self):
self.ReportServiceStatus(win32service.SERVICE_START_PENDING)
try:
self.log('START: Service start')
self.ReportServiceStatus(win32service.SERVICE_RUNNING)
self.start()
win32event.WaitForSingleObject(self.stop_event, win32event.INFINITE)
except Exception as e:
self.log('Exception: {}'.format(e))
self.SvcStop()
def SvcStop(self):
self.log('STOP: Service stopping...')
self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
self.stop()
win32event.SetEvent(self.stop_event)
self.ReportServiceStatus(win32service.SERVICE_STOPPED)
def log(self, msg):
servicemanager.LogInfoMsg(str(msg)) #system log
logging.info(str(msg)) #text log
def start(self):
self.runflag = True
while self.runflag:
win32api.Sleep((2*1000), True)
self.log('Service alive')
def stop(self):
self.runflag = False
self.log('Stop received')
if __name__ == '__main__':
win32serviceutil.HandleCommandLine(MyWindowsService)
In the script I use a log file to check if it's working properly. I'm running python3.6 (also tried with python3.4) on Windows 7 and I'm experiencing the following problem. When I run python MyWindowsService.py install the prompt says that the service has been installed (but nothing is written in the log file). If I try to start the service, I get Service Error: 1 - More info NET HELPMSG 3547 which doesn't say much about the error. If I run python MyWindowsService.py debug, the program runs just fine (the log file is written), but still I don't have any control over the service: if I open another prompt and try to stop/start the service I still got the same results as stated above.
I also tryed to insert some debug code inside the init function, and when I run python MyWindowsService.py install it seems it doesn't get called. Is it possible?
I've checked for multiple solution and workarounds around the net, but I didn't find anything suitable. What am I missing?

As pointed out by eriksun in the comment to the first post, the problem came from the location of the python script, that was in a drive mapped with an UNC path - I'm working with a virtual machine. Moving the python script in the same drive as the python installation did the job.
To sum it up for future uses, if the service fails to start and you're pretty sure about your code, these are helpful actions to try and solve your issues:
use sc start ServiceName, sc query ServiceName and sc stop ServiceName to get info about the service.
check if your file is in a physical drive or in a UNC-mapped drive. If the latter try to run the script using the UNC path (for example python \\Server\share\python\your-folder\script.py) or move your script in the same drive as the python installation
make sure that "python36.dll", "vcruntime140.dll", and "pywintypes36.dll" are either symlink'd to the directory that has PythonService.exe; or symlink'd to the System32 directory; or that the directories with these DLLs are in the system (not user) Path
Check the system register with command reg query HKLM\System\CurrentControlSet\Services\your_service_name /s to get more information about the script
PLease, feel free to complete, change, modify the last so that it can be usefull for anyone that like me encounder this issue.
EDIT: One more thing... My project was thought to work actually with network folders (and UNC-mapped drive) and it failed when I tried to make it run as service. One very useful (day-saving) resource that I used to make it work is the SysinternalsSuite by Mark Russinovich that I found in this post. Hope this helps.

How can I make a Python package which serves a website with Flask to work with Apache?

I have written a Python package hwrt (see installation instructions if you want to try it) which serves a website when executed with
$ hwrt serve
2014-12-04 20:27:07,182 INFO * Running on http://127.0.0.1:5000/
2014-12-04 20:27:07,183 INFO * Restarting with reloader
I would like to let it run on http://www.pythonanywhere.com, but when I start it there I get
19:19 ~ $ hwrt serve
2014-12-04 19:19:59,282 INFO * Running on http://127.0.0.1:5000/
Traceback (most recent call last):
File "/home/MartinThoma/.local/bin/hwrt", line 108, in <module>
main(args)
File "/home/MartinThoma/.local/bin/hwrt", line 102, in main
serve.main()
File "/home/MartinThoma/.local/lib/python2.7/site-packages/hwrt/serve.py", line 95, in main
app.run()
File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 739, in run
run_simple(host, port, self, **options)
File "/usr/local/lib/python2.7/dist-packages/werkzeug/serving.py", line 613, in run_simple
test_socket.bind((hostname, port))
File "/usr/lib/python2.7/socket.py", line 224, in meth
return getattr(self._sock,name)(*args)
socket.error: [Errno 98] Address already in use
I only found this in the documentation:
Flask
never use app.run(), it will break your webapp. Just import the
app into your wsgi file...
By searching for wsgi file, I found mod_wsgi (Apache). However, I don't understand how I can adjust my current minimalistic Flask application to work with that. Currently, the script behind hwrt serve is:
#!/usr/bin/env python
# -*- coding: utf-8 -*-
"""Start a webserver which can record the data and work as a classifier."""
import pkg_resources
from flask import Flask, request, render_template
from flask_bootstrap import Bootstrap
import os
import json
# hwrt modules
import hwrt
import hwrt.utils as utils
def show_results(results, n=10):
"""Show the TOP n results of a classification."""
import nntoolkit
classification = nntoolkit.evaluate.show_results(results, n)
return "<pre>" + classification.replace("\n", "<br/>") + "</pre>"
# configuration
DEBUG = True
template_path = pkg_resources.resource_filename('hwrt', 'templates/')
# create our little application :)
app = Flask(__name__, template_folder=template_path)
Bootstrap(app)
app.config.from_object(__name__)
#app.route('/', methods=['POST', 'GET'])
def show_entries():
heartbeat = request.args.get('heartbeat', '')
return heartbeat
#app.route('/interactive', methods=['POST', 'GET'])
def interactive():
if request.method == 'POST':
raw_data_json = request.form['drawnJSON']
# TODO: Check recording
# TODO: Submit recorded json to database
# Classify
model_path = pkg_resources.resource_filename('hwrt', 'misc/')
model = os.path.join(model_path, "model.tar")
print(model)
results = utils.evaluate_model_single_recording(model, raw_data_json)
# Show classification page
page = show_results(results, n=10)
page += 'back'
return page
else:
# Page where the user can enter a recording
return render_template('canvas.html')
def get_json_result(results, n=10):
s = []
for res in results[:min(len(results), n)]:
s.append({res['semantics']: res['probability']})
return json.dumps(s)
#app.route('/worker', methods=['POST', 'GET'])
def worker():
# Test with
# wget --post-data 'classify=%5B%5B%7B%22x%22%3A334%2C%22y%22%3A407%2C%22time%22%3A1417704378719%7D%5D%5D' http://127.0.0.1:5000/worker
if request.method == 'POST':
raw_data_json = request.form['classify']
# TODO: Check recording
# TODO: Submit recorded json to database
# Classify
model_path = pkg_resources.resource_filename('hwrt', 'misc/')
model = os.path.join(model_path, "model.tar")
results = utils.evaluate_model_single_recording(model, raw_data_json)
return get_json_result(results, n=10)
else:
# Page where the user can enter a recording
return "Classification Worker (Version %s)" % hwrt.__version__
def get_parser():
"""Return the parser object for this script."""
from argparse import ArgumentParser, ArgumentDefaultsHelpFormatter
parser = ArgumentParser(description=__doc__,
formatter_class=ArgumentDefaultsHelpFormatter)
return parser
def main():
app.run()
if __name__ == '__main__':
main()

Ok, a not so non-sequitur answer to your question is around what mod_wsgi does to interface with your app. A typical flask app would look something like this:
from flask import Flask
app = Flask(__name__)
app.route("/")
def hello():
return "Holy moly that tunnel was bright.. said Bit to NIC"
if __name__ == "__main__":
app.run()
Unfortunately, Apache has no way to know what to do with this (though the app would run happily on its own). In order to get the app and Apache to play nice together we're going to use something called mod_wsgi. What Mod_WSGI does that's important to us, is that it provides a known interface (a file type called wsgi) that's going to wrap our application and initialize it so that we can serve it through Apache.
I'm going to assume you are using a python virtual environment, but if you aren't you can omit the step that deals with this in the instructions below. If you're curious why virtual environments are so great, feel free read about the python ecosystem.
Also - you can include an extra flag (assuming you are running wsgi as a daemon) to automatically reload the daemon whenever you touch or alter your wsgi file. This is quite useful during development and debugging so I'll include is below.
Anyway, let's get started. I'll break this down to steps below.
Configuring Apache for mod_wsgi
Enable mod_wsgi in Apache:
sudo apt-get install libapache2-mod-wsgi
Edit your /etc/apache2/sites-available/<yoursite>.conf.
<VirtualHost interface:port>
WSGIDaemonProcess yourapp user=someUser processes=2 threads=15
WSGIProcessGroup yourapp
# In this case / refers to whatever relative URL path hosts flask
WSGIScriptAlias / /absolute/path/to/yourapp.wsgi
<Directory /path/to/your/main/py/file/ >
# Use good judgement here when server hardening, this assumes dev env
Order allow,deny
Allow from all
Require all granted
#The below enables 'auto-reload' of WSGI
WSGIScriptReloading On
</Directory>
# If you want to serve static files as well and bypass flask in those cases
Alias /relative/url/to/static/content/
<Directory /absolute/path/to/static/root/directory/>
Order allow,deny
Allow from all
</Directory>
</VirtualHost>
Create your yourapp.wsgi file and put it in the appropriate place: Be wary of file permissions!
#!/usr/bin/python
import sys
import logging
# Activate virtual environment.
# If you are not using venv, skip this.
# But you really should be using it!
activate_this = "/path/to/venv/bin/activate_this.py"
execfile(activate_this, dict(__file__=activate_this))
# Handle logging
logging.basicConfig(stream=sys.stderr)
sys.path.insert(0, "/path/to/your/main/py/file/")
from YourMainPyFileName import app as application
application.secret_key = "your_secret_key"
Reload Apache and troubleshoot problems. I set this up probably every few weeks for a different project or idea I have and... I usually have to fix one thing or another when doing it from scratch. Don't despair though! Flask has great documentation on this.
Once you've done all this you should be at a place where flask runs all on its own. The sample flask app above is the actual code I use to verify everything works whenever I set this up.
This was left here in case it's some use, but is not really directly related to the question...
The answer here is to use x-send-file. This takes advantage of letting Apache do what it's good at (serving static content), while at the same time first letting flask (or other python framework) do it's work first. I do this often to let flask handle my auth layers in single page web apps and have so far been happy with the results.
Doing so requires two things:
First - Enable xsendfile on Apache2 sudo apt-get install libapache2-mod-xsendfile.
Second - Alter your apache2 configuration so allow x-send-file headers:
Alter your conf file in /etc/apache2/sites-available/<yoursite>.conf and add...
XSendFile On
XSendFilePath /path/to/static/directory
This can be entered top level within the <Virtualhost></Virtualhost> tags.
Don't forget to restart Apache sudo service apache2 restart.
Finally - Configure your flask app to use x-send-file in your app.py file:
app.user_x_sendfile = True
Note: Must be done after app initialization. Consequently can also be passed as an initialization parameter.
Flask has documentation on this (excerpt below):
use_x_sendfile
Enable this if you want to use the X-Sendfile feature. Keep in mind that the server has to support this. This only affects files sent with the send_file() method.
New in version 0.2.
This attribute can also be configured from the config with the USE_X_SENDFILE configuration key. Defaults to False.

I ran into a similar issue #moose was having. Getting connection refused and couldnt even telnet localhost 5000.
Turns out theres a ports.conf file i had to add Listen 5000
Happy days.

Using Python 3.3 in OpenShift's book example

OpenShift recently published a book, "Getting Started with OpenShift". It is a good guide for someone just starting out.
In Chapter 3 they show how to modify a template application to use Python 2.7 and Flask. Our requirement is for Python 3.3.
On Page 19, one of the modifications to wsgi.py is: execfile(virtualenv, dict(file=virtualenv)). execfile was done away with in 3.x. There are examples in StackOverflow on how to translate but it is not clear to me how to apply those to this case.
Does anyone have any insight into this issue?

As indicated in this question, you can replace the line
execfile(virtualenv, dict(__file__=virtualenv))
by
exec(compile(open(virtualenv, 'rb').read(), virtualenv, 'exec'), dict(__file__=virtualenv))
In my opinion, it would be better to break this up into a few simpler pieces. Also we should use a context handler for the file handling::
with open(virtualenv, 'rb') as exec_file:
file_contents = exec_file.read()
compiled_code = compile(file_contents, virtualenv, 'exec')
exec_namespace = dict(__file__=virtualenv)
exec(compiled_code, exec_namespace)
Breaking it up in this way will also make debugging easier (actually: possible). I haven't tested this but it should work.

If you are facing issues with virtualenv settings on wsgi.py file in Python3 I have solved just deleting it.
This is my wsgi.py file and it's working
#!/usr/bin/python
from flaskapp import app as application
if __name__ == '__main__':
from wsgiref.simple_server import make_server
httpd = make_server('0.0.0.0', 5000, application)
# Wait for a single request, serve it and quit.
#httpd.handle_request()
httpd.serve_forever()

Easy application logging/debugging with nginx, uwsgi, flask?

I'm not looking to turn on the dangerous debugging console, but my application is getting a 500 error and doesn't seem to be writing any output for me to investigate more deeply.
I saw this exchange on the mailing list, which led me to this page on logging errors.
However, I still find this very confusing and have a couple of questions:
(1) In which file should the stuff below go?
ADMINS = ['yourname#example.com']
if not app.debug:
import logging
from logging.handlers import SMTPHandler
mail_handler = SMTPHandler('127.0.0.1',
'server-error#example.com',
ADMINS, 'YourApplication Failed')
mail_handler.setLevel(logging.ERROR)
app.logger.addHandler(mail_handler)
...assuming the "getting bigger" file pattern for larger applications? __init__.py? config.py? run.py?
(2) I am overwhelmed by options there, and can't tell which I should use. Which loggers should I turn on, with what settings, to replicate the local python server debug I get to stdout when I run run.py? I find that default, local output stream very useful, more so than the interactive debugger in the page. Does anyone have a pattern they could share on setting up something replicating this with an nginx deployment, outputting to a log?
(3) Is there anything I need to change, not at the flask level, but in nginx, say in my /etc/nginx/sites-available/appname file, to enable logging?
UPDATE
Specifically, I'm looking for information like I get when python runs locally as to why, say, a package isn't working, or where my syntax error might be, or what variable doesn't exist:
$ python run.py
Traceback (most recent call last):
File "run.py", line 1, in <module>
from myappname import app
File "/home/me/myappname/myappname/__init__.py", line 27, in <module>
file_handler.setLevel(logging.debug)
File "/usr/lib/python2.7/logging/__init__.py", line 710, in setLevel
self.level = _checkLevel(level)
File "/usr/lib/python2.7/logging/__init__.py", line 190, in _checkLevel
raise TypeError("Level not an integer or a valid string: %r" % level)
When I run flask on a server, I never see this. I just get a uWSGI error in the browser, and have no idea which code was problematic. I would just like something like the above to be written to a file.
I notice also that setting the following logging didn't really write much to file, even when I turn the log way up to the DEBUG level:
from logging import FileHandler
file_handler = FileHandler('mylog.log')
file_handler.setLevel(logging.DEBUG)
app.logger.addHandler(file_handler)
mylog.log is blank, even when my application errors out.
I'll also add that I've tried to set debug = True in the following ways, in __init__.py:
app = Flask(__name__)
app.debug = True
app.config['DEBUG'] = True
from werkzeug.debug import DebuggedApplication
app.wsgi_app = DebuggedApplication(app.wsgi_app, True)
app.config.from_object('config')
app.config.update(DEBUG=True)
app.config['DEBUG'] = True
if __name__ == '__main__':
app.run(debug=True)
While in my config.py file, I have...
debug = True
Debug = True
DEBUG = True
Yet, no debugging happens, and without logging or debugging, this is rather hard to track down. Errors simply terminate the application with the un-useful browser message:
uWSGI Error
Python application not found

Set config['PROPAGATE_EXCEPTIONS'] to True when running app in production and you want tracebacks to be logged into log files. (I haven't tried with SMTP handler, though..)

The part where you create handlers, add to loggers etc. should be in the if __name__ == '__main__' clause, i.e. your main entry point. I assume that would be run.py.
I'm not sure I can answer this - it depends on what you want. I'd advise looking at the logging tutorial to see the various options available.
I don't believe you need to change anything at the nginx level.
Update: You might want to have an exception clause that covers uncaught exceptions, e.g.
if __name__ == '__main__':
try:
app.run(debug=True)
except Exception:
app.logger.exception('Failed')
which should write the traceback of any exception which occurred in app.run() to the log.

I know that this is a VERY old post, but I ran into the issue now, and it took me a bit to find the solution. Flask sends errors to the server. I was running Gunicorn with an upstart script on Ubuntu 14.04 LTS, and the place where I found the error logs was as follows:
/var/log/upstart/myapp.log
http://docs.gunicorn.org/en/stable/deploy.html#upstart
Just in case some other poor soul ends up in this situation.

How do I deploy web2py on PythonAnywhere?

How do i get a basic web2py server up and running on
PythonAnywhere?

[update - 29/05] We now have a big button on the web tab that will do all this stuff for you. Just click where it says Web2Py, fill in your admin password, and you're good to go.
Here's the old stuff for historical interest...
I'm a PythonAnywhere developer. We're not massive web2py experts (yet?) but I've managed to get web2py up and running like this:
First download and unpack web2py:
wget http://www.web2py.com/examples/static/web2py_src.zip
unzip web2py_src.zip
Go to the PythonAnywhere "Web" panel and edit your wsgi.py. Add these lines:
import os
import sys
path = '/home/my_username/web2py'
if path not in sys.path:
sys.path.append(path)
from wsgihandler import application
replacing my_username with your username.
You will also need to comment out the last two lines in wsgi.py, where we have the default hello world web.py application...
# comment out these two lines if you want to use another framework
#app = web.application(urls, globals())
#application = app.wsgifunc()
Thanks to Juan Martinez for his instructions on this part, which you can view here:
http://web2py.pythonanywhere.com/
then open a Bash console, and cd into the main web2py folder, then run
python web2py.py --port=80
enter admin password
press ctrl-c
(this will generate the parameters_80.py config file)
then go to your Web panel on PythonAnywhere, click reload web app,
and things should work!

You can also simply run this bash script:
http://pastebin.com/zcA5A89k
admin will be disabled because of no HTTPS unless you bypass it as in the previous post. It will create a security vulnerability.

Pastebin was down, I retrieved this from the cache.
cd ~
wget -O web2py_srz.zip http://web2py.com/examples/static/web2py_src.zip
unzip web2py_src.zip
echo "
PATH = '/home/"`whoami`"/web2py'
import os
import sys
sys.stdout = sys.stderr
os.chdir(PATH)
if not './' in sys.path[:1]: sys.path.insert(0,'./')
from gluon.main import wsgibase as application
" > /var/www/wsgi.py
cd web2py
python -c "from gluon.main import save_password; save_password(raw_input('admin password: '),433)"

I have recently summarized my experience with deployment of Web2Py on PythonAnywhere here
Hope it helps
NeoToren

I'll try to add something new to the discussion. The EASIEST way I've found is to go here when you aren't logged in. This makes it so you don't have to mess around with the terminal:
https://www.pythonanywhere.com/try-web2py
Come up with a domain name, then you'll get redirected to a page showing your login information and created dashboard for that domain. From there just create an account so your app isn't erased after 24 hours. When you sign up, your app has a 3 month expiry date (if you're not paying). I believe this is a new policy. Then simply go to https://appname.pythonanywhere.com/admin and then enter the password you were given and then upload your Web2Py file into the dashboard and then visit the page.
I'm not sure how to upload a Web2Py app on PythonAnywhere for an existing account, but that's the easiest method I've found.