Hello friends, I am using flask-pymongo
app = Flask(name)
app.config['MONGO_URI'] = 'mongodb://localhost:27017/shiptest'
mongo = PyMongo(app)
if name == 'main': app.run() #this line causing error
I am running the file in virtual environment.
If I write app.run(), I get error:
Debug mode: off Permission denied.
If I write app.run(debug=True), I get error:
Debug mode: on Permission denied,
If I write app.run(debug=False), I get error:
Debug mode: off Permission denied,
Please help to resolve the error.
nkahr
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