I created some custom classifiers locally and then i try to deploy on bluemix an app that classifies an image based on the classifiers i made.
When I try to deploy it, it failes to start.
import os
import json
from os.path import join, dirname
from os import environ
from watson_developer_cloud import VisualRecognitionV3
import time
start_time = time.time()
visual_recognition = VisualRecognitionV3(VisualRecognitionV3.latest_version, api_key='*************')
with open(join(dirname(__file__), './test170.jpg'), 'rb') as image_file:
print(json.dumps(visual_recognition.classify(images_file=image_file,threshold=0, classifier_ids=['Angle_971786581']), indent=2))
print("--- %s seconds ---" % (time.time() - start_time))
Even if I try to deploy a simple print , it failes to deploy, but the starter app i get from bluemix, or a Flask tutorial (https://www.ibm.com/blogs/bluemix/2015/03/simple-hello-world-python-app-using-flask/) i found online deploy just fine.
I'm very new to web programming and using cloud services so i'm totally lost.
Thank you.
Bluemix is expecting your python application to serve on a port. If your application isn't serving some kind of response on the port, it assumes the application failed to start.
# On Bluemix, get the port number from the environment variable PORT
# When running this app on the local machine, default the port to 8080
port = int(os.getenv('PORT', 8080))
#app.route('/')
def hello_world():
return 'Hello World! I am running on port ' + str(port)
if __name__ == '__main__':
app.run(host='0.0.0.0', port=port)
It looks like you're writing your code to just execute once and stop. Instead, make it do the work when someone hits your URL, like shown in the hello_world() function above.
Think about what you want to happen when someone goes to YOUR_APP_NAME.mybluemix.net
If you do not want your application to be a WEB application, but instead just execute once (a background worker application), then use the --no-route option at the end of your cf push command. Then, look at the logs using cf logs appname --recent to see the output of your application
https://console.ng.bluemix.net/docs/manageapps/depapps.html#deployingapps
The main problem was watson-developer-cloud module, giving me an error that it could not be found.
I downgraded to python version 2.7.12, installing it for all users.
Modified runtime.exe and requirments.txt (requirments.txt possible not needed)
Staged with Diego, using no-route and set-health-check APP_NAME none command.
Those fixed the problem, but i still get an exit status 0.
when you deploy an app in bluemix,you should have a requirements.txt which include services you used in your app.
so ,you should checkout your requirements.txt,maybe you lost
watson_developer_cloud
and then, the requirements.txt likes this:
Flask==0.10.1
watson_developer_cloud
Related
I've uploaded some code into a server. The code was working locally but when I upload it to the server it gives me an Internal Server Error. The website is running with wsgi and the code is:
try:
from decksite import main, APP as application
except Exception as e:
from shared import repo
repo.create_issue('Error starting website', exception=e)
if __name__ == '__main__':
print('Running manually. Is something wrong?')
application.run(host='0.0.0.0', debug=False)
So both the try and the except are failing. I want to add a second exception and pass it all to a simple flask application that would output both exceptions to the browser and log them to a file. The problem is that I don't know how to pass the exception to the error_app flask app and that it breaks in the line where I set the logging config. Here is what I've done. I'm only getting NoneType: None instead of the full exception.
import os, sys
sys.path.append("/home/myuser/public_html/flask");
try:
from decksite import main, APP as application
except Exception as error:
#from shared import repo
#repo.create_issue('Error starting decksite', exception=error)
#sys.path.insert(0, os.path.dirname(__file__))
#from error_app import app as application
# This is the code that goes into the error flask application
import logging
import traceback
from flask import Flask, __version__
app = Flask(__name__)
application = app
#app.route("/")
def hello():
return traceback.format_exc()
# The next line gives Internal Server Error
logging.basicConfig(filename='example.log', level=logging.DEBUG)
logging.exception(error)
return traceback.format_exc()
if __name__ == '__main__':
print('Running manually. Is something wrong?')
application.run(host='0.0.0.0', debug=False)
I don't have sudo in the server and can't ssh to it so unless I'm able to log the errors I'm not going to be able to fix anything.
Edit: I've almost got it as I want:
.htaccess
website.wsgi
error_app.py
website/init.py
website/main.py
Create a custom 500 handler and print out the trackback
import traceback
#app.errorhandler(500)
def internal_server_error(e):
return render_template('500_error.html', traceback=traceback.format_exc())
Have your '500_error.html' template show you the traceback.
You mentioned 500 Internal Server Error is coming. Things are proper in your local but fail on server. Since you don't have ssh access, it might be tough to debug. If you use something like Docker or Kubernetes to build and deploy it can be useful. I can suggest you some ways to debug. The code is not going to try except and failing, the possible reason is the server itself not starting due to missing requirements say some import or it could be anything.
Debug Steps
Create a virtual environment and re-install requirements in it similar to your server. This will help you to identify if there is a missing requirement.
if you are environment is not production and you are testing the application in the server then put debug = True. It will show an error on the front end. This is definitely not a recommended approach. I am suggesting since you don't have ssh access.
If possible create a simple route say /hello and in that just return hello, and check whether it is returning you the right result or not. This will let you know whether your server is starting or not. Even this is not recommendable for production.
You can also check the flask app before request and after request. This might also be useful
Hopefully, 1st step debugging will help you a lot and you might not need to go to step 2 and step 3. I gave you for the worst-case scenario
I've started learning Flask to develop web applications. What I am realy missing is an automatic Browser refresh after any code change (including static files, templates, etc.). This seems to be a standard feature in almost any Javascript framework. Frontend people have several terms for that: auto reload / refresh, hot reload / refresh (hotreload), live reload / refresh (livereload), ...
Here on Stackoverflow most similar questions are related to the auto-reload of the Flask server (--> https://stackoverflow.com/search?q=flask+auto+reload).
J just want a simple browser refresh.
I googled and tried several things - with no luck:
https://github.com/lepture/python-livereload
https://gist.github.com/amontalenti/8922609
How can I have a smooth developing experience with Flask without having to hit the F5 key 1000 times a day in a browser just to see the results of my changes?
I think the answer is somewhere near to python-livereload from the link above.
So I guess an alternative title of my question could be:
Does somebody have a working example of Flask + python-livereload?
I am to dumb to get it from their documentation :)
EDIT: for the sake of completenes here is the Flask app I am using.
# filename: main.py
from flask import Flask, render_template
from livereload import Server
app = Flask(__name__)
#app.route('/')
def index():
return "INDEX"
#app.route('/bart')
def use_jinja():
return render_template('basic.html')
if __name__ == '__main__':
server = Server(app.wsgi_app)
server.serve(port=5555)
I start the app with
python main.py
This is an interesting question you've raised so I built a quick and dirty Flask application which utilizes the livereload library. Listed below are the key steps for getting this to work:
Download and install the livereload library:
pip install livereload
Within your main file which starts up your flask application, run.py in my particular case, wrap the flask application with the Server class provided by livereload.
For example, my run.py file looks like the following:
from app import app
from livereload import Server
if __name__ == '__main__':
server = Server(app.wsgi_app)
server.serve()
Start your server again:
python run.py
Navigate to localhost in the browser and your code changes will be auto-refreshed. For me I took the default port of 5500 provided by livereload so my url looks like the following: http://localhost:5500/.
With those steps you should now be able to take advantage of auto-reloads for your python development, similar to what webpack provides for most frontend frameworks.
For completeness the codebase can be found here
Hopefully that helps!
If on Windows:
set FLASK_ENV=development
I use Guard::LiveReload becose solution with python-livereload don't work correct with debug (for me).
In first terminal emulator:
$ cd my-flask-project
$ flask run
* Serving Flask app 'webface' (lazy loading)
* Environment: development
* Debug mode: on
* Running on http://127.0.0.1:54321/ (Press CTRL+C to quit)
* Restarting with stat
* Debugger is active!
In secondu terminal emulator:
$ cd my-flask-project
$ guard init livereload
Now I edit Guardfile and add/edit watch. Like this:
watch(%r{.+\.py})
watch(%r{webface/.+\.(css|js|html(\.j2)?)})
again in second terminal emulator:
$ guard
17:29:25 - INFO - LiveReload is waiting for a browser to connect.
17:29:25 - INFO - Guard is now watching at '/home/marek/my-flask-project'
[1] guard(main)> 17:29:29 - INFO - Browser connected.
17:29:31 - INFO - Reloading browser: webface/templates/base.html.j2
17:45:41 - INFO - Reloading browser: webface/templates/base.html.j2
[1] guard(main)>
Maybe need to click livereload extention button in browser to connect browser to guard.
I'm creating a GroupMe bot that's hosted on Heroku with Gunicorn as the WSGI server.
When I try deploying the app, I get a failed to find object 'app' in 'MODULE_NAME' error, I think because I don't have a WSGI callable.
Here's what I have:
def app():
while True:
rqResponse = requests.get('https://api.groupme.com/v3/groups/' + groupID +'/messages', params = requestParams)
# Pings the gm-membot Heroku app so it doesn't idle.
requests.get('http://gm-bot.herokuapp.com')
if rqResponse.status_code == 200:
gotten = rqResponse.json()['response']['messages']
for message in gotten:
messageText = message['text'].lower()
if (messageText in bot_reply.staticTriggers) or (messageText in bot_reply.dynamicTriggers):
bot_reply.botReply(message)
requestParams['since_id'] = message['id']
else:
raise Exception('error')
break
time.sleep(5)
My Procfile output:
web: gunicorn MODULE_NAME:app --workers=1
However, After looking at the documentation for Gunicorn and WSGI, I can't figure out how to mesh it with the code I already have written using the Requests library. Is there any way I can get Gunicorn to work without a lot of rewriting? Also, I'm very new to this, so I apologize if there's an obvious answer.
(P.S. everything works fine if I just host the app on my latptop!)
I found two answers: first and second, though I think it's the memory leak on server (cause you said hosted local everything works).
Try and let me know
I have a python app developed on Flask. Everything works fine offline, I tried deploying on CherryPy successfully too. Now, I'm trying to deploy the same on www.pythonanywhere.com.
Here's the deploy.py I use for deploying the Flask app on CherryPy
from cherrypy import wsgiserver
from appname import app
def initiate():
app_list = wsgiserver.WSGIPathInfoDispatcher({'/appname': app})
server = wsgiserver.CherryPyWSGIServer( ('http://username.pythonanywhere.com/'), app_list)
try:
server.start()
except KeyboardInterrupt:
server.stop()
print "Server initiated..."
initiate()
print "Ended"
I created a "manual configuration" app on pythonanywhere.com.
Here's the configuration file (username_pythonanywhere_com_wsgi.py):
import sys
path = '/home/username/appname'
if path not in sys.path:
sys.path.append(path)
import deploy
deploy.initiate()
Now I'm pretty sure that it "almost worked", because in the server logs I could see my "Server initiated..." message.
2013-09-27 09:57:16 +0000 username.pythonanywhere.com - *** Operational MODE: single process ***
Server initiated...
Now the problem, when I try to view my app username.pyhtonanywhere.com/about, it times out.
This I believe is caused due to incorrect port given while starting the CherryPy server (in deploy.py).
Could anyone please tell how I can properly initiate the CherryPy server?
Joe Doherty is right. You want something more like this in you wsgi file:
import sys
sys.path = [ <path to your web app> ] + sys.path
from cherrypy._cpwsgi import CPWSGIApp
from cherrypy._cptree import Application
from <your_web_app> import <your web app class>
config_path = '<path to your cherrypy config>'
application = CPWSGIApp(
Application(<your web app class>(), '', config = config_path)
I stuck everything that should be based on your particular app in <>s.
I have a flask app where I'd like to execute some code on the first time the app is run, not on the automatic reloads triggered by the debug mode. Is there any way of detecting when a reload is triggered so that I can do this?
To give an example, I might want to open a web browser every time I run the app from sublime text, but not when I subsequently edit the files, like so:
import webbrowser
if __name__ == '__main__':
webbrowser.open('http://localhost:5000')
app.run(host='localhost', port=5000, debug=True)
You can set an environment variable.
import os
if 'WERKZEUG_LOADED' in os.environ:
print 'Reloading...'
else:
print 'Starting...'
os.environ['WERKZEUG_LOADED']='TRUE'
I still don't know how to persist a reference that survives the reloading, though.
What about using Flask-Script to kick off a process before you start your server? Something like this (cribbed from their documentation and edited slightly):
# run_devserver.py
import webbrowser
from flask.ext.script import Manager
from myapp import app
manager = Manager(app)
if __name__ == "__main__":
webbrowser.open('http://localhost:5000')
manager.run(host='localhost', port=5000, debug=True)
I have a Flask app where it's not really practical to change the DEBUG flag or disable reloading, and the app is spun up in a more complex way than just flask run.
#osa's solution didn't work for me with flask debug on, because it doesn't have enough finesse to pick out the werkzeug watcher process from the worker process that gets reloaded.
I have this code in my main package's __init__.py (the package that defines the flask app). This code is run by another small module which has from <the_package_name> import app followed by app.run(debug=True, host='0.0.0.0', port=5000). Therefore this code is executed before the app starts.
import ptvsd
import os
my_pid = os.getpid()
if os.environ.get('PPID') == str(os.getppid()):
logger.debug('Reloading...')
logger.debug(f"Current process ID: {my_pid}")
try:
port = 5678
ptvsd.enable_attach(address=('0.0.0.0', port))
logger.debug(f'========================== PTVSD waiting on port {port} ==========================')
# ptvsd.wait_for_attach() # Not necessary for my app; YMMV
except Exception as ex:
logger.debug(f'PTVSD raised {ex}')
else:
logger.debug('Starting...')
os.environ['PPID'] = str(my_pid)
logger.debug(f"First process ID: {my_pid}")
NB: note the difference between os.getpid() and os.getppid() (the latter gets the parent process's ID).
I can attach at any point and it works great, even if the app has reloaded already before I attach. I can detach and re-attach. The debugger survives a reload.