Flask maintenance mode

app.py

from flask import Flask, redirect, url_for, request

app = Flask(__name__)

is_maintenance_mode = True

# Always throw a 503 during maintenance: http://is.gd/DksGDm

@app.before_request
def check_for_maintenance():
    if is_maintenance_mode and request.path != url_for('maintenance'): 
        return redirect(url_for('maintenance'))
        # Or alternatively, dont redirect 
        # return 'Sorry, off for maintenance!', 503


@app.route('/')
def index():
    return 'Hello!'

@app.route('/maintenance')
def maintenance():
    return 'Sorry, off for maintenance!', 503

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

Nice solution!, thanks :)

"if is_maintenance_mode and request.path != url_for('maintenance') and not 'static' in request.path "

With that, the site still showing all the styles and does not show the site as plain text

Best regards and thanks again

Better solution IMHO:

from flask import Flask, abort
import os

app = Flask(__name__)

@app.before_request
def check_under_maintenance():
    if os.path.exists("maintenance"): # Check if a "maintenance" file exists (whatever it is empty or not)
        abort(503) # No need to worry about the current URL, redirection, etc

@app.route('/')
def index():
    return "This is Admiral Ackbar, over"

@app.errorhandler(503)
def error_503(error):
    return "It's a trap!", 503

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

@app.before_request
def check_for_maintenance():
    rule = request.url_rule.rule
    if is_maintenance_mode and 'maintenance' not in rule: 
        return redirect(url_for('maintenance'))

Also works