I want to implement simple site layout:
/ must render home.html
/one, /two, /three must render one.html, two.html, three.html correspondingly
So far I came up with following code:
main_page = Blueprint('main', __name__)
category_page = Blueprint('category', __name__)
#main_page.route("/")
def home():
return render_template('home.html')
#category_page.route('/<category>')
def show(category):
return render_template('{}.html'.format(category))
app = Flask(__name__)
app.register_blueprint(main_page, url_prefix='/')
app.register_blueprint(category_page, url_prefix='/categories')
This way I am able to route categories to /categories/<category>. How can I route them to just /<category> instead, while keeping home.html linked to /? Appreciate your help
I tried two approaches:
Setting url_prefix='/' for both blueprints => second one does not work.
Instead of main_page blueprint use just app.route('/') to render home.html. This one also does not work when mixing with category_page blueprint
You can move the variable to the url_prefix parameter inside the registering statement :
#main_page.route("/")
def home():
return render_template('home.html')
app.register_blueprint(main_page, url_prefix='/')
#category_page.route('/')
def show(category):
return render_template('{}.html'.format(category))
app.register_blueprint(category_page, url_prefix='/<category>')
(it depends on the complexity of the whole pattern, but it may be better to keep the registering statements with the variables close to each function to handle many views.)
Related
How can I decouple this class? I would like to put the paths in another file, is it possible to move the routes in another file?
#api.route('/home', '/api/email')
class Server(Resource):
def create_server(app, oauth=None):
if not oauth:
oauth = default_provider(app)
app = prepare_app(app)
#app.before_request
def load_current_user():
user = User.query.get(1)
g.user = user
#app.route('/home')
def home():
return 'home'
#app.route('/oauth/authorize', methods=['GET', 'POST'])
#oauth.authorize_handler
def authorize(*args, **kwargs):
return True
Those
#app.route('/home') # and
#app.route('/oauth/authorize', methods=['GET', 'POST'])
have to be in another file.
My attempt was this, I tried to create a file for routers:
class Router():
def __init__(self, app, oauth):
self.app = app
self.oauth = oauth
#app.route('/home')
def home():
return 'home'
I'm getting this error:
NameError: name 'app' is not defined
Well, I see a package you can use for flask projects called Flask-Via [pypi], inspired by the Django URL configuration system and designed to add similar functionality to Flask applications that have grown beyond a simple single file application. The following example is given from the docs of this project:
from flask import Flask
from flask.ext.via import Via
from flask.ext.via.routers.default import Functional
app = Flask(__name__)
def foo(bar=None):
return 'Foo View!'
routes = [
Functional('/foo', foo),
Functional('/foo/<bar>', foo, endpoint='foo2'),
]
via = Via()
via.init_app(app, route_module='flask_via.examples.basic')
if __name__ == "__main__":
app.run(debug=True)
I think this is exactly what you want :) and you can move it to another python module for example called routers.py.
I would like to be able to generate a navbar based on an object's contents, basically having a navbar array or class that holds subpages and generates the appropriate navbar with collapsing parts etc. I have experience with Laravel in PHP which is similar to Flask but I can't figure out a way to do it easily. I would have to provide a set of data objects to every single page since it's part of the layout but would prefer not to have to specify it specifically for each page. Is there a way to do this?
So far I only have the basics, an app factory, view and blueprint:
Factory
def create_app():
app = Flask(__name__)
app.config.from_pyfile('config.py')
app.register_blueprint(blueprint_index)
return app
Blueprint
from flask import Blueprint, render_template, session
blueprint_index = Blueprint('simple_page', __name__)
#blueprint_index.route('/')
def index():
if 'text' in session:
session['text'] += 1
else:
session['text'] = 0
return render_template('pages/index.html', text=str(session['text']))
Ignore the little bit of debug text I added to the route.
You could do this with a context processor. Add this to the end of your blueprint code-block:
#blueprint_index.context_processor
def inject_text():
return dict(text=session['text'])
{{text}} will now be available within every template on that blueprint.
I am building a Flask app with a blueprint mounted on two different endpoint (one is a legacy alias to the other).
In my blueprint class:
ldp = Blueprint('ldp', __name__)
#ldp.route('/<path:uuid>', methods=['GET'])
#ldp.route('/', defaults={'uuid': None}, methods=['GET'],
strict_slashes=False)
def get_resource(uuid):
# Route code...
In my main server code:
app = Flask(__name__)
app.config.update(config['flask'])
app.register_blueprint(ldp, url_prefix='/new_ep')
# Legacy endpoint. #TODO Deprecate.
app.register_blueprint(ldp, url_prefix='/old_ep')
How can I get the actual URL of the request up to the /old_ep or /new_ep part in the route method, e.g. http://localhost:5000/new_ep?
So far I have used
request.host_url + request.path.split('/')[1]
but it looks quite inelegant and possibly error-prone. I would like to use the information from the blueprint setup if possible.
Thanks for your help.
EDIT: I could get to the Blueprint instance from within the request with
current_app.blueprints[request.blueprint]
and I was hoping that the url_prefix attribute that I set when registering the blueprint was there, but it is None instead. As I read from the documentation for the supposedly related iter_blueprints() method, apparently these blueprints are listed without regard of how many times and with which parameters they were registered. Too bad.
Here is a full working example to get the idea based off issue 612
from flask import Flask, Blueprint, url_for, request, g
bp = Blueprint('whatever', __name__)
#bp.url_defaults
def bp_url_defaults(endpoint, values):
url_prefix = getattr(g, 'url_prefix', None)
if url_prefix is not None:
values.setdefault('url_prefix', url_prefix)
#bp.url_value_preprocessor
def bp_url_value_preprocessor(endpoint, values):
g.url_prefix = values.pop('url_prefix')
#bp.route('/something')
def index():
return 'host prefix is %s%s' % (request.host_url, g.url_prefix)
app = Flask(__name__)
app.register_blueprint(bp, url_prefix='/new_ep', url_defaults={'url_prefix': 'new_ep'})
app.register_blueprint(bp, url_prefix='/old_ep', url_defaults={'url_prefix': 'old_ep'})
I'm building the Flask app with React, I ended up having a problem with routing.
The backend is responsible to be an API, hence some routes look like:
#app.route('/api/v1/do-something/', methods=["GET"])
def do_something():
return something()
and the main route which leads to the React:
#app.route('/')
def index():
return render_template('index.html')
I'm using react-router in the React app, everything works fine, react-router takes me to /something and I get the rendered view, but when I refresh the page on /something then Flask app takes care of this call and I get Not Found error.
What is the best solution? I was thinking about redirecting all calls which are not calling /api/v1/... to / it's not ideal as I will get back the home page of my app, not rendered React view.
We used catch-all URLs for this.
from flask import Flask
app = Flask(__name__)
#app.route('/', defaults={'path': ''})
#app.route('/<path:path>')
def catch_all(path):
return 'You want path: %s' % path
if __name__ == '__main__':
app.run()
You can also go an extra mile and reuse the Flask routing system to match path to the same routes as client so you can embed the data client will need as JSON inside the HTML response.
Maybe as extension to the answers before. This solved the problem for me:
from flask import send_from_directory
#app.route('/', defaults={'path': ''})
#app.route('/<path:path>')
def serve(path):
path_dir = os.path.abspath("../build") #path react build
if path != "" and os.path.exists(os.path.join(path_dir, path)):
return send_from_directory(os.path.join(path_dir), path)
else:
return send_from_directory(os.path.join(path_dir),'index.html')
For some reason, the catch-all URLs did not work for me. I found that using the flask 404 handler results in the exact same thing. It sees the url and passes it down to react where your router will handle it.
#app.errorhandler(404)
def not_found(e):
return app.send_static_file('index.html')
Just to inform handle error 404 and render_template works perfectly for me.
#app.errorhandler(404)
def not_found(e):
return render_template("index.html")
I have to combine both catch-all and 404 handler for it to work properly. I am hosting a react-app in a subpath with its own redirection handler from react-router.
#app.route('/sub-path', defaults={'path': 'index.html'})
#app.route('/sub-path/<path:path>')
def index(path):
return send_from_directory('../react-dir/build', path)
#app.errorhandler(404)
def not_found(e):
return send_from_directory('../react-dir/build','index.html')
I'm using Flask 0.8.
How to have an alias of a URL like this:
#app.route('/')
def index():
# I want to display as http://localhost/index, BUT, I DON'T WANT TO REDIRECT.
# KEEP URL with only '/'
#app.route('/index')
def index():
# Real processing to display /index view
So, why my hope to use an alias because of DRY of processing /index
Someone knew the solution?
thanks pepperists.
This should work. But why do you want two URL's to display the same thing?
#app.route('/')
#app.route('/index')
def index():
...
As is written in URL registry doc of Flask :
You can also define multiple rules for the same function. They have to
be unique however.
#app.route('/users/', defaults={'page': 1})
#app.route('/users/page/<int:page>')
def show_users(page):
pass
I don't know if Flask has a way to assign more than one URL to a view function, but you could certainly chain them like this:
#app.route('/')
def root():
return index()
#app.route('/index')
def index():
# Real processing to display /index view