How can I activate the Flask debugger when running under mod_wsgi?
I have DEBUG, PROPAGATE_EXCEPTION and PRESERVE_CONTEXT_ON_EXCEPTION set to True, but still the debugger doesn't appear on exceptions.
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2 Answers
As described in the Flask documentation at:
http://flask.pocoo.org/docs/quickstart/#debug-mode
use:
app.debug = True
Under mod_wsgi you aren't doing the app.run() though.
Ensure you are setting 'app.debug' at global scope and not in a conditional section where checking whether __name__ is __main__.
answered May 12, 2012 at 23:15
Graham Dumpleton
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5 Comments
jd.
I just put
app.debug = True in my .wsgi file and it still doesn't work. What I found that works is from werkzeug.debug import DebuggedApplication application = DebuggedApplication(app, True).
Graham Dumpleton
That is the long handed way of achieving the same. Can't argue with you since you don't actually provide what you actually have in your WSGI file to see what you did when setting app.debug.
kerma
From flask.pocoo.org/docs/quickstart/#debug-mode "Even though the interactive debugger does not work in forking environments (which makes it nearly impossible to use on production servers), it still allows the execution of arbitrary code. This makes it a major security risk and therefore it must never be used on production machines." Which to me means that app.debug=True should not work under mod_wsgi by default
Graham Dumpleton
Not all execution modes of mod_wsgi are multi process. The default for daemon mode is a single process where the debugger will work fine.
Yuvi
Running my script as CGI on Apache, @jd's solution worked for me while normal app.debug did not
You may use interactive debugger provided by werkzeug:
from werkzeug.debug import DebuggedApplication
application = DebuggedApplication(app, True)
This is earley suggested by @jd work for me.
1 Comment
Vitold S.
@DanChaltiel after create app variable i.e. after code like app = Flask()