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If the Head First Python quick_session.py (flask) code is good, is something wrong with my environment?

I copied this test code from the book and also downloaded it from the book's site. It runs and when I enter "http://localhost:5000/setuser/Fred", it replies "User value set to: Fred", but when I enter "http://localhost:5000/getuser", there is a 404 error.
Since I see no errors in the code, is there something in my Python environment that is causing this? Here is the short bit of code. Thanks from a beginner.
from flask import Flask, session
app = Flask(__name__)
app.secret_key = 'YouWillNeverGuess'
#app.route('/setuser/<user>')
def setuser(user: str) -> str:
session['user'] = user
return 'User value set to: ' + session['user']
#app.route('/getuser')
def getuser() -> str:
return 'User value is currently set to: ' + session['user']
if __name__ == '__main__':
app.run(debug=True)
* Serving Flask app 'quick_session' (lazy loading)
* Environment: development
* Debug mode: on
* Running on http://127.0.0.1:5000 (Press CTRL+C to quit)
* Restarting with stat
* Debugger is active!
* Debugger PIN: 474-778-940

It turns out I needed to restart Windows. Then the code ran as expected.
Back in the old days we sang the tune "FDISK Format Reinstall Doo-Dah Doo-Dah"... Same song, newer beat.

Flask gives 404 when trying to render a template

#app.route('/')
def index():
return render_template("home.html")
My folder structure looks like this
tree-/
-static/
-styles.css
-templates/
-home.html
-app.py
I get
Not Found
The requested URL was not found on the server. If you entered the URL manually please check your spelling and try again.
On the browser and
127.0.0.1 - - [01/May/2021 09:41:47] "GET / HTTP/1.1" 404 -
In the debugger
Have looked at other related posts saying stuff about trailing slashes and it doesn't look like its making a difference, either I access
http://127.0.0.1:5000
or
http://127.0.0.1:5000/
I run my application using
app = Flask(__name__)
if __name__ == '__main__':
app.run()
I get
* Serving Flask app "app" (lazy loading)
* Environment: production
WARNING: This is a development server. Do not use it in a production deployment.
Use a production WSGI server instead.
* Debug mode: off
* Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)
from flask import Flask, redirect, url_for, render_template
app = Flask(__name__)
if __name__ == '__main__':
app.run()
#app.route('/')
def index():
return "Hello World"

Problem is order of code.
Line app.run() has to be at the end of code because it runs endless code/loop (which gets requests from clients/browsers and sends reponses) and it blocks next lines of code. All code after app.run() is executed after server is closed.
from flask import Flask
app = Flask(__name__)
#app.route('/')
def index():
return "Hello World"
if __name__ == '__main__':
app.run()

Move app.run() to the bottom.
Thanks to furas for this solution, hope this can help people looking at this thread

How do you customize the launching message for Flask?

I'm running a Flask app in Cloud9. Whenever I start my Flask app, it says this message:
* Serving Flask app "app" (lazy loading)
* Environment: production
WARNING: This is a development server. Do not use it in a production deployment.
Use a production WSGI server instead.
* Debug mode: off
* Running on http://0.0.0.0:80/ (Press CTRL+C to quit)
Is there a way to change this message? I'd like it to say something like this:
Connect to me at http://0.0.0.0:80/!
I've searched stack overflow and the web but couldn't find anything. I'm starting my app with app.run().
Also, is it possible to make the URL cyan?

You can change everything besides Running on http://0.0.0.0:80/ (Press CTRL+C to quit) by changing show_server_banner of flask.cli:
from flask import Flask
import sys
cli = sys.modules['flask.cli']
# put your own message here
cli.show_server_banner = lambda *x: click.echo("My nice message")
app = Flask(__name__)
app.run(host='0.0.0.0', port='80')
To get rid of the Running on http://0.0.0.0:80/ ... message, you can use unittest.mock:
from unittest import mock
from werkzeug._internal import _log
def my_startup_log(*args):
# log all messages except for the * Running on message
if not args[1].startswith(" * Running on"):
return _log(*args)
app = Flask(__name__)
with mock.patch('werkzeug.serving._log') as mocked:
# patch the logger object and replace with own logger
mocked.side_effect = my_startup_logger
app.run(host='0.0.0.0', port='8000')
This is very hacky and depends on the internal implementation of flask. Be careful when using this in production code, as this could easily break.

Posting to Flask app gives 404 even though route is defined

I ran the following Flask app and tried to post to /, but I got a 404 response. The / route is defined, why doesn't it get handled?
from flask import Flask, request
app = Flask(__name__)
app.run()
#app.route('/', methods=['POST'])
def handle_post_request():
return str(request.form)
C:\Python35\python.exe D:/Dropbox/dev/contact.py
* Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)
127.0.0.1 - - [03/Apr/2017 20:46:01] "POST / HTTP/1.1" 404 -

Calling app.run blocks, no code is executed after it until the server is stopped. Make sure your app is completely set up before calling run. In this case, move the route definition above the call to run.
Flask 0.11 introduced a more robust way to start applications to avoid this and other common mistakes. Completely remove app.run(). Set the FLASK_APP environment variable to your app's entry point, and call flask run.
FLASK_APP=contact.py flask run
You can also set FLASK_DEBUG=1 to enable debug mode.
In Windows, use set FLASK_APP=contact.py to set environment variables instead.

Unable to open web browser link after starting flask server

My twilio_receive_msg.py file bears the following code:
from flask import Flask, request, redirect
import twilio.twiml
app = Flask(__name__)
#app.route("/", methods=['GET', 'POST'])
def hello_monkey():
"""Respond to incoming calls with a simple text message."""
resp = twilio.twiml.Response()
resp.message("Hello, Mobile Monkey")
return str(resp)
if __name__ == "__main__":
app.run(debug=True)
When I run python twilio_receive_msg.py, the code keeps running like forever and I see the following.
* Running on http://127.0.0.1:5000/ (Press CTRL+C to quit)
* Restarting with stat
* Debugger is active!
* Debugger pin code: 135-602-230
Also, I am not able to open 'http://127.0.0.1:5000/'. I paste the link in my browser, a message appears that says waiting for 127.0.0.1.... and this message persists even after a long wait!
My OS is Windows and browser is chrome. I tried Explorer too but got the same message.
Can anybody assist me with this? Why am I not able to open the above link?

After long searches i tried localhost instead of 127.0.0.1 in the internet explorer and it worked.