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understanding the decorators and need to return inside wrapper function

This is the code from api.py
def route(self,path):
if path in self.routes :
raise AssertionError('Such route already exists')
def wrapper(handler):
self.routes[path] =handler
print(colorama.Fore.GREEN,"handler check",handler)
return handler
return wrapper
And this the code from the app.py
from api import API
app = API(templates_dir="templates")
from middleware import Middleware
#app.route("/home")
def home (request,response):
response.text ='Hello from home function'
#app.route("/about")
def about(request,response):
response.text ='Hello from about function'
#app.route("/hello/{name}")
def greeting(request, response, name):
response.text = f"Hello, {name}"
I am having difficulty in understanding how route decorator works? I see that it takes in a parameter "/about" and "/home" but why do we need to return the reference to the handler in the route decorator.
A clear explanation should really help.

So route is in a class that looks something like this:
class API:
routes = {}
def route(self, path):
if path in self.routes :
raise AssertionError('Such route already exists')
def wrapper(handler):
self.routes[path] =handler
print(colorama.Fore.GREEN,"handler check",handler)
return handler
return wrapper
def handle(self, path, request):
if path in self.routes:
response = Response()
self.routes[path](request, response)
send_to_client(response)
raise NotFound()
When you use
#app.route("/home")
def home(request, response):
response.text ='Hello from home function'
this is the same as doing something like:
def home_function(request, response):
response.text ='Hello from home function'
home = app.route("/home")(home_function)
First we're calling app.route with the argument "/home". First thing this does is check if we already have a route called /home. Then it returns the new function, the one that's called wrapper.
Now we call wrapper with the function we want to handle the /home path with. What this does is add to the dict a mapping from the path (/home) to the function that we want to call to handle this path.
The same thing happens for all decorated functions.
Now when we want to handle an incoming request, you can imaging something a bit like the handle function runs. We know that every key of the routes mapping is going to be a path, and every value is going to be a function that takes a request and a response. So then we just need to get the right function, and call the it appropriately.
Does that make sense?

How do you break Python Flask controller flow when calling return inside a function?

I'm using Python Flask and I got defined the following function:
def verify_session():
if not 'logged_in' in session:
flash("You are not logged in.<br/>Please, log in to use our application.", "warning")
return redirect(url_for('login_path'))
This is meant to be called in every controller. I know that I could use #app.before_request but on some places I dont want this function to be called (in the same Login page, for example) and I really dont want the function to check the excluded paths.
My problem is this: inside any Flask controller for each of the paths the first lines look like this:
#app.route('/web/account', methods=["GET"], endpoint="account_path")
def account():
verify_session()
return render_template('account')
But, despite I get the "You need to login" flashed message, it still returns the Account page, because the return inside verify_session just returns for this function.
What I need is something like return verify_session() but that doesn't end the controller flow unless it actually is necessary to return it.
Maybe this problem is only Python/Programming specific rather than Flask so I excuse myself and ask moderators to edit my Question accordingly. I mention Flask because it's the enviroment where I'm having the trouble.
IF more information is needed I'm willing to update. Thank you in advance.

result = verify_session()
if not result:
result = render_template("account")
return result
but really you should be using a decorator for this
def verify_session(fn):
def inner(*args,**kwargs):
if not 'logged_in' in session:
flash("You are not logged in.<br/>Please, log in to use our application.", "warning")
return redirect(url_for('login_path'))
return fn(*args,**kwargs)
return inner
then you would use it like
#app.route('/web/account', methods=["GET"], endpoint="account_path")
#verify_session #make sure to use any decorators after the route decorator
def account():
return render_template('account')
lastly ... I hope you are aware that you are reinventing the wheel and you should probably just use flask-login to manage this stuff(unless you have a pretty compelling reason not to)

To add to Joran's excellent answer. Instead of using #app.before_request, you can put your login required routes in a blueprint and then use #blueprint.before_request:
account = Blueprint('account', __name__)
#account.before_request
def verify_session():
...
#account.route('/')
def index():
...
Then in your app:
app = Flask(__name__)
app.register_blueprint(account, url_prefix='/web/account')

Python AppEngine, how to take parameters to another page

In my app, each page is each Python class. From page A, I want to take some data in this page and redirect to page B.
Does Google App Engine has some ways to do it ? (I don't want to use something like: global variable or cookie for this small work)
Thanks :)

You can compute the value you need in the first handler (AHandler), then redirect to the second handler (BHandler) passing that value as a GET parameter. Finally BHandler, reads that parameter and does something with it. Here is some code:
import urllib
class AHandler(webapp2.RequestHandler):
def get(self):
name = 'Some name'
redirect('b?%s' % urllib.urlencode({'name': name}))
class BHandler(webapp2.RequestHandler):
def get(self):
name = self.request.get('name')
# do something with name

Calling another view in Pyramid

My goal: In Pyramid, to call another view-callable, and to get a Response object back without knowing any details about that view-callable.
In my Pyramid application, say I have a view "foo" which is defined using a view_config decorator:
#view_config(route_name="foo",
renderer="foo.jinja2")
def foo_view(request):
return {"whereami" : "foo!"}
Now say that I want to route "bar" to a view that does the same thing for the time being, so it internally calls foo_view and returns its Response:
#view_config(route_name="bar")
def bar_view(request):
return foo_view(request)
...but wait! That doesn't work, since foo_view doesn't return a Response, its renderer does.
So, this will work:
#view_config(route_name="bar",
renderer="foo.jinja2")
def bar_view(request):
return foo_view(request)
as it will apply the same renderer as foo_view did. But this is bad, as I now must repeat myself by copying the renderer value AND having to know the renderer of the view being called.
So, I am going to hope that there is some function available in Pyramid that allows calling another view-callable and getting a Response object back without knowing or caring how it was rendered:
#view_config(route_name="bar")
def bar_view(request):
response = some_function_that_renders_a_view_callable(foo_view, request)
return response
What would some_function_that_renders_a_view_callable be?
pyramid.views.render_view appears to search for a view by name; I don't want to give my views names.
(Note: Returning HTTPFound to cause the client to redirect to the target route is what I am trying avoid. I want to "internally" redirect).

Yep. There is some concerns
doesn't return a Response
predicates/renderer
permissions
request properties associated to old request
Thats why you should not call view from view as function, unless you know what you doing
Pyramid creators did awesome tool for server side redirect - http://docs.pylonsproject.org/projects/pyramid/en/latest/narr/subrequest.html

You can invoking a view with using request.invoke_subrequest:
from wsgiref.simple_server import make_server
from pyramid.config import Configurator
from pyramid.request import Request
def view_one(request):
subreq = Request.blank('/view_two')
response = request.invoke_subrequest(subreq)
return response
def view_two(request):
request.response.body = 'This came from view_two'
return request.response
if __name__ == '__main__':
config = Configurator()
config.add_route('one', '/view_one')
config.add_route('two', '/view_two')
config.add_view(view_one, route_name='one')
config.add_view(view_two, route_name='two')
app = config.make_wsgi_app()
server = make_server('0.0.0.0', 8080, app)
server.serve_forever()`
When /view_one is visted in a browser, the text printed in the
browser pane will be "This came from view_two". The view_one view
used the pyramid.request.Request.invoke_subrequest() API to obtain a
response from another view (view_two) within the same application
when it executed. It did so by constructing a new request that had a
URL that it knew would match the view_two view registration, and
passed that new request along to
pyramid.request.Request.invoke_subrequest(). The view_two view
callable was invoked, and it returned a response. The view_one view
callable then simply returned the response it obtained from the
view_two view callable.

I was struggling with this as well. I have a solution using the render_to_response method, though I'm sure there's a "more correct" way to do it. Until someone posts it, however, here is how I handled this:
from pyramid.renderers import render_to_response
#view_config(route_name="foo", renderer="foo.mak")
def foo_view(request):
return {'stuff':'things', '_renderer':'foo.mak')
def bar_view(request):
values = foo_view(request)
renderer = values['_renderer']
return render_to_response(renderer,values)
(Pyramid 1.3)
This requires a renderer to be used, but by declaring that renderer in the original view's return values, you can retrieve it in another view without knowing what it is. I'm suspecting the need to do this isn't easily findable because there's other, better methods for accomplishing tasks solved by this solution.
Another shortcoming is that it relies on direct import of the view callable. It would be nice if it could be looked up directly by route.

The Pyramid documentation here indicates that leaving the name key word argument out of view_config will cause the view to be registered by the function itself (rather than a string):
Such a registration... implies that the view name will be *my_view*
So, in your case you should be able to use pyramid.view.render_view or pyramid.view.render_view_to_response referencing foo_view directly:
#view_config(route_name="bar")
def bar_view(request):
return pyramid.views.render_view_to_response(None, request, name=foo_view)
Update:
Yep, your right, passing the view function does not work.
It's interesting, but taking your example code and applying the route_name to the config
did not work for me. However, the following example, just giving the view a name sets the route url
and gives the view a name. In this fashion render_view_to_response works as advertised. Naming,
your views may not be what you want, but this configuration accomplishes the same thing as your
example code without added configuration.
#view_config(name="foo")
def foo_view(request):
# returning a response here, in lieu of having
# declared a renderer to delegate to...
return Response('Where am i? `{0[whereami]}'.format({"whereami" : "foo!"}))
#view_config(name="bar")
def bar_view(request):
# handles the response if bar_view has a renderer
return render_view_to_response(None, request, name='foo')
#view_config(name="baz")
def baz_view(request):
# presumably this would not work if foo_view was
# not returning a Response object directly, as it
# skips over the rendering part. I think you would
# have to declare a renderer on this view in that case.
return foo_view(request)
if __name__ == '__main__':
config = Configurator()
config.scan()
app = config.make_wsgi_app()
serve(app, host='127.0.0.1', port='5000')

Not the precise solution you asked for, but a solution to the problem you describe:
Create a view class, of which both foo and bar are methods. Then bar can call self.foo()
Common view_configuration, such as the template name can be applied to the class, and then you can decorate each method with just the view name.
In short, the following should meet your needs, if I understand the problem correctly.
#view_defaults(renderer="foo.jinja2")
class WhereaboutsAreFoo(object):
#view_config(route-name="foo")
def foo_view(self):
return {"whereami" : "foo!"}
#view_config(route-name="bar")
def bar_view(self):
return self.foo_view()

can't you do something like that:
#view_config(name="baz")
def baz_view(request):
return HTTPFound(location=self.request.route_path('foo'))

using a colon in a string in python

i am using python 2.6.5 to develop an app for google app engine - i am not too familiar with python, but i'm learning.
i am trying to put a url into a string so variable = "string http://domain.name"
then i print the string out. the problem is, if the colon (after http) is in the string, i don't get any output and i don't know why.
i've tried escaping the string with:
"""http://domain.name"""
r"http://domain.name"
"http\://domain.name"
"http\://domain.name"
"http\\://domain.name"
"http:://domain.name"
none of them seem to work and i'm not sure what else to try
The context is like so
variables.py is:
...
HOST_URL = "http://domain.name"
...
example logout.py
import variables
import sys
...
class Logout(webapp.RequestHandler):
""" RequestHandler for when a user wishes to logout from the system."""
def post(self):
self.get()
def get(self):
print(variables.HOST_URL)
print('hi')
self.redirect(variables.HOST_URL)
sys.exit()
or
in file functions.py
import variables
import sys
...
def sendhome(requesthandler)
print 'go to '+variables.HOST_URL
requesthandler.redirect(variables.HOST_URL)
sys.exit()
called from a context like:
from functions import sendhome
...
class Logout(webapp.RequestHandler):
""" RequestHandler for when a user wishes to logout from the system."""
def post(self):
self.get()
def get(self):
sendhome(self)
any help would be appreciated
thanks

If I'm not terrible mistaken, GAE uses WSGI, you do not simply print things, you are supposed to return a proper HTTP response object (it is not PHP).
I guess that if you access the page using firefox+firebug and look at the network->header you will see that the browser is taking http: as an HTTP header with value "//domain.name".
Edited: By the way, should not you be using "self.response.out.write" instead of "print"?

The problem was the sys.exit() after the call to print or redirect