Flask App Implementing Trouble .htaccess

[centellajs]

Really Thx in advanced .

 

Hi i am just new in the python world , i need to execute a flaskapp for whatsapp sending messages, i followed the steps to implement the flaskapp but i can't execute any code for python native or flask worst it produces an error i followed all the steps to implement flask

 

 

I really need to make it works the .htaccess looks like

 

/home/xtremese/public_html/inteygratemyfrank-i/.htaccess

RewriteEngine On

RewriteBase /

RewriteRule ^(media/.*)$ - [L]

RewriteRule ^(admin_media/.*)$ - [L]

RewriteRule ^(inteygratemyfrank-i\.wsgi/.*)$ - [L]

RewriteRule ^(.*)$ inteygratemyfrank-i/flask.wsgi/$1 [QSA,PT,L]

 

 

my flask.wsgi looks like

 

import os, sys

# edit your username below

sys.path.append("/home/xtremese/public_html/inteygratemyfrank-i");

sys.path.insert(0, os.path.dirname(__file__))

from myapp import app as application

# make the secret code a little better

application.secret_key = 'secret'

Create a python script named myapp.py

import sys

from flask import Flask, __version__

app = Flask(__name__)

application = app

@app.route("/")

def hello():

return """

HelioHost rules!<br><br>

<a href="/flask/python/version/">Python version</a><br>

<a href="/flask/flask/version/">Flask version</a>

"""

@app.route("/python/version/")

def p_version():

return "Python version %s" % sys.version

@app.route("/flask/version/")

def f_version():

return "Flask version %s" % __version__

if __name__ == "__main__":

app.run()

 

and my files structure looks like :

 

[Krydos]

As it says at the top of http://wiki.helionet.org/Flask flask isn't available on Johnny. If you want to use flask with python 2.7 you need to be on the Ricky server, and if you want to use flask with python 3.6 you need to be on the Tommy server. Let us know if you need help moving.