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pythonapacheflaskmod-wsgiwerkzeug

Why isn't Flask giving me an interactive debugger?

Here's a simple Flask app with a syntax error (it should be sys.version, not sys.version()

from flask import Flask, url_for, request, render_template
import sys

app = Flask(__name__)
app.config.from_object(__name__)

@app.route('/')
def index():
    version = sys.version()
    return "This is the index page.  Python version info: {}".format(version)

if __name__ == '__main__':
    app.run(debug=True)

When I run it locally and try to access http://localhost:5000, I get the nice Werkzeug debugger:

enter image description here

but when I run it on a remote server (Apache 2.4.17, mod_wsgi 4.4.21), I only get a generic 500 error page:

enter image description here

I've tried various things to make this work:

  • set app.debug = True right after app = Flask(__name__)
  • based on this link, I tried adding this code as well:

from werkzeug.debug import DebuggedApplication application = DebuggedApplication(app, True)

but nothing seems to help.

I know that the debugger doesn't work in forked environments, but according to this link, the default mode for mod_wsgi is to not be forked. My Apache config is:

WSGIDaemonProcess flasktest user=david group=devgrp threads=1 python-path=/apps/www/80/wsgi-scripts/FlaskTest:/apps/.virtualenvs/flasktest/lib/python3.4/site-packages

So why can't I get the pretty debugger?

EDIT: Here's the modified file with my additions (that aren't working)

from flask import Flask
import sys

app = Flask(__name__)
app.config.from_object(__name__)
app.debug = True

from werkzeug.debug import DebuggedApplication
application = DebuggedApplication(app, True)  

@app.route('/')
def index():   
    version = sys.version()
    return "This is the index page.  Python version info: {}".format(version)

if __name__ == '__main__':
    app.run(debug=True)
Solution

  • To add WSGI middleware to a Flask app, wrap and replace the app's wsgi_app attribute. In the case of the debug middleware, set app.debug = True as well so that Flask doesn't convert exceptions to 500 responses.

    app.debug = True
app.wsgi_app = DebuggedApplication(app.wsgi_app, evalex=True)

    It is a security risk to run the debugger on a public server. Despite the recent change requiring a PIN before running code, the debugger may still be insecure in unforseen ways. Typically, you get information about errors in production by using logging. This answer is a good basic summary of how to set that up.

    Your method would probably work if you had replaced app rather than calling it application. mod_wsgi was probably picking up app before application, so it was still running the original app. It's still recommended to wrap app.wsgi_app though.