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I writing a flask, trying to organize it by using Blueprint with Namespace, following this tutorial
I had faced some problem, and had look around internet and had review solution in 1 and 2. The first one is not relevant to what I doing, and the second one the solution just doesn't fix my problem.
Here are my code:
project/project.py
from flask import Flask, jsonify, url_for
from .apis.apis import api
app = Flask(__name__)
app.register_blueprint(api, url_prefix="/api")
project/apis/apis.py
from flask import Blueprint
from .user.authentication import auth
from flask_restplus import Api, apidoc, Resource
blueprint = Blueprint("api", __name__)
api = Api(blueprint, doc='/docs', ui=False)
api.add_namespace(auth, path="/auth") #Keep getting error at this line
project/apis/user/authentication.py
from flask_restplus import Namespace
auth = Namespace('auth', description='Authentication')
#auth.route("/test")
def authentication():
return "test"
Stack Trace
Traceback (most recent call last):
File "/home/gaara/Python/Flask-Api/project/__init__.py", line 1, in <module>
from .project import app
File "/home/gaara/Python/Flask-Api/project/project.py", line 3, in <module>
from .apis.apis import api
File "/home/gaara/Python/Flask-Api/project/apis/apis.py", line 13, in <module>
api.add_namespace(auth, path="/auth")
File "/home/gaara/Python/Flask-Api/venv/lib/python3.6/site-packages/flask_restplus/api.py", line 413, in add_namespace
self.register_resource(ns, resource, *self.ns_urls(ns, urls), **kwargs)
File "/home/gaara/Python/Flask-Api/venv/lib/python3.6/site-packages/flask_restplus/api.py", line 255, in register_resource
self._register_view(self.app, resource, *urls, **kwargs)
File "/home/gaara/Python/Flask-Api/venv/lib/python3.6/site-packages/flask_restplus/api.py", line 276, in _register_view
resource_func = self.output(resource.as_view(endpoint, self, *resource_class_args,
AttributeError: 'function' object has no attribute 'as_view'
I am not sure why I keep getting this error, had try few approach, include put apis.py all in __init__.py and change the import, but always getting the same error.
What I wish is to code api in an organize way, and when go to localhost:5000/api/auth/test it will output me test
You defined a function, but Flask restplus requires a class, as you can also see in your tutorial.
So it should look like this:
from flask_restplus import Resource
#auth.route("/test")
class Authentication(Resource):
def get(self):
return "test"
While the approved answer is fully correct, let me dispose why we always say, that if something is possible, does not mean that you should do!
So the following to be for everyone a lesson (at 1st place for myself) of an ANTI-PATTERN!
I was a bit tinkering here & there, I found out what is needed for a function like you mentioned in order to be registered and usable by a framework, like F-Restplus (I've used in my case F-RestX==0.5.0, but in reality is a fork of that), and here we go:
def avg(dummy_self_var=''):
# the dummy_self_var is needed in order to trick and pretend to be as an instance method
result = 'any-calculation-that-has-to-be-made'
logging.debug(f'AVG result is: {result}')
return Response(str(result))
# imitating that this function is a Resource :D
avg.view_class = Resource
avg.as_view = Resource.as_view # at the end, it could be the following function as well: `lambda x,y: avg`
avg.view_class.methods = {'GET', }
avg.view_class.get = avg
api.add_resource(avg, '/asd')
With the help of this I achieved to have the same functionality, it works &
gets registered automatically by the Swagger-UI docs:
So, while nearly everything is possible, I could not imagine a situation where someone needs this 'workaround', instead, to be forward-compatible, I would definitely refactor instead of producing this mess in the long-run. Of course, the choice is up to you.
One of the libraries my project requires that a folder with the CSS files that were in the application root called "themes". web.py by default, uses the folder "static" to return the static file and just rename her... not One of the solutions I found online was the following
in urls it is necessary to add the line
'/(?:img|js|css)/.*', 'app.controllers.public.public',
in app.controllers.public
require nex code
class public:
def GET(self):
public_dir = 'themes'
try:
file_name = web.ctx.path.split('/')[-1]
web.header('Content-type', mime_type(file_name))
return open(public_dir + web.ctx.path, 'rb').read()
except IOError:
raise web.notfound()
def mime_type(filename):
return mimetypes.guess_type(filename)[0] or 'application/octet-stream'
but this solution does not work and files are still picked up from static...
is there a simple and clear solution to the problem? maybe we should change the name of the folder inside the web.py?
There's no simple way to change web.py's use of /static/, but there is a really easy way to add one of your own, with no need to add anything to your list of urls.
Look at web.py's code and you'll find web.httpserver.StaticMiddleware is where this is defined. Your job, create another WSGI middleware, with the new prefix. Then, because this is WSGI middleware, add your new class to the run chain.
from web.httpserver import StaticMiddleware
if __name__ == '__main__':
app = web.application(urls, globals())
app.run(lambda app: StaticMiddleware(app, '/themes/')
If that was too terse for you, consider it's the same as explicitly creating a new subclass and passing that subclass to app.run():
from web.httpserver import StaticMiddleware
class MyStaticMiddleware(StaticMiddleware):
def __init__(self, app, prefix='/themes/'):
StaticMiddleware.__init__(self, app, prefix)
if __name__ == '__main__':
app = web.application(urls, globals())
app.run(MyStaticMiddleware)
Note that '/static/' will still work, loading files from the /static/ subdirectory: All you've done is added another processor, which does the same thing, but from the '/themes/' subdirectory.
I'm writing test cases for code that is called via a route under Flask. I don't want to test the code by setting up a test app and calling a URL that hits the route, I want to call the function directly. To make this work I need to mock flask.request and I can't seem to manage it. Google / stackoverflow searches lead to a lot of answers that show how to set up a test application which again is not what I want to do.
The code would look something like this.
somefile.py
-----------
from flask import request
def method_called_from_route():
data = request.values
# do something with data here
test_somefile.py
----------------
import unittest
import somefile
class SomefileTestCase(unittest.TestCase):
#patch('somefile.request')
def test_method_called_from_route(self, mock_request):
# want to mock the request.values here
I'm having two issues.
(1) Patching the request as I've sketched out above does not work. I get an error similar to "AttributeError: 'Blueprint' object has no attribute 'somefile'"
(2) I don't know how to exactly mock the request object if I could patch it. It doesn't really have a return_value since it isn't a function.
Again I can't find any examples on how to do this so I felt a new question was acceptable.
Try this
test_somefile.py
import unittest
import somefile
import mock
class SomefileTestCase(unittest.TestCase):
def test_method_called_from_route(self):
m = mock.MagicMock()
m.values = "MyData"
with mock.patch("somefile.request", m):
somefile.method_called_from_route()
unittest.main()
somefile.py
from flask import request
def method_called_from_route():
data = request.values
assert(data == "MyData")
This is going to mock the entire request object.
If you want to mock only request.values while keeping all others intact, this would not work.
A few years after the question was asked, but this is how I solved this with python 3.9 (other proposed solutions stopped working with python 3.8 see here). I'm using pytest and pytest-mock, but the idea should be the same across testing frameworks, as long as you are using the native unittest.mock.patch in some capacity (pytest-mock essentially just wraps these methods in an easier to use api). Unfortunately, it does require that you set up a test app, however, you do not need to go through the process of using test_client, and can just invoke the function directly.
This can be easily handled by using the Application Factory Design Pattern, and injecting application config. Then, just use the created app's .test_request_context as a context manager to mock out the request object. using .test_request_context as a context manager, gives everything called within the context access to the request object. Here's an example below.
import pytest
from app import create_app
#pytest.fixture
def request_context():
"""create the app and return the request context as a fixture
so that this process does not need to be repeated in each test
"""
app = create_app('module.with.TestingConfig')
return app.test_request_context
def test_something_that_requires_global_request_object(mocker, request_context):
"""do the test thing"""
with request_context():
# mocker.patch is just pytest-mock's way of using unittest.mock.patch
mock_request = mocker.patch('path.to.where.request.is.used')
# make your mocks and stubs
mock_request.headers = {'content-type': 'application/json'}
mock_request.get_json.return_value = {'some': 'json'}
# now you can do whatever you need, using mock_request, and you do not
# need to remain within the request_context context manager
run_the_function()
mock_request.get_json.assert_called_once()
assert 1 == 1
# etc.
pytest is great because it allows you to easily setup fixtures for your tests as described above, but you could do essentially the same thing with UnitTest's setUp instance methods. Happy to provide an example for the Application Factory design pattern, or more context, if necessary!
with help of Gabrielbertouinataa on this article: https://medium.com/#vladbezden/how-to-mock-flask-request-object-in-python-fdbc249de504:
code:
def print_request_data():
print(flask.request.data)
test:
flask_app = flask.Flask('test_flask_app')
with flask_app.test_request_context() as mock_context:
mock_context.request.data = "request_data"
mock_context.request.path = "request_path"
print_request_data()
Here is an example of how I dealt with it:
test_common.py module
import pytest
import flask
def test_user_name(mocker):
# GIVEN: user is provided in the request.headers
given_user_name = "Some_User"
request_mock = mocker.patch.object(flask, "request")
request_mock.headers.get.return_value = given_user_name
# WHEN: request.header.get method is called
result = common.user_name()
# THEN: user name should be returned
request_mock.headers.get.assert_called_once_with("USERNAME", "Invalid User")
assert result == given_user_name
common.py module
import flask
def user_name():
return flask.request.headers.get("USERNAME", "Invalid User")
What you're trying to do is counterproductive. Following the RFC 2616 a request is:
A request message from a client to a server includes, within the first line of that message, the method to be applied to the resource, the identifier of the resource, and the protocol version in use.
Mocking the Flask request you need to rebuild its structure, what certainly, you will not to want to do!
The best approach should be use something like Flask-Testing or use some recipes like this, and then, test your method.
In a pyramid app I am building (called pyplay), I need to retrieve an application setting that I have in development.ini. The problem is that the place where I am trying to get that setting cannot access the request variable (e.g. at the top level of a module file).
So, after looking at this example in the documentation: http://docs.pylonsproject.org/projects/pyramid_cookbook/en/latest/configuration/django_settings.html I started doing something very simple and hardcoded at first just to make it work.
Since my development.ini has this section: [app:main], then the simple example I tried is as follows:
from paste.deploy.loadwsgi import appconfig
config = appconfig('config:development.ini', 'main', relative_to='.')
but the application refuses to start and displays the following error:
ImportError: <module 'pyplay' from '/home/pish/projects/pyplay/__init__.pyc'> has no 'main' attribute
So, thinking that maybe I should put 'pyplay' instead of 'main', I went ahead, but I get this error instead:
LookupError: No section 'pyplay' (prefixed by 'app' or 'application' or 'composite' or 'composit' or 'pipeline' or 'filter-app') found in config ./development.ini
At this point I am a bit stuck and I don't know what am I doing wrong. Can someone please give me a hand on how to do this?
Thanks in advance!
EDIT: The following are the contents of my development.ini file (note that pish.theparam is the setting I am trying to get):
###
# app configuration
# http://docs.pylonsproject.org/projects/pyramid/en/latest/narr/environment.html
###
[app:main]
use = egg:pyplay
pyramid.reload_templates = true
pyramid.debug_authorization = false
pyramid.debug_notfound = false
pyramid.debug_routematch = false
pyramid.default_locale_name = en_US.utf8
pyramid.includes =
pyramid_debugtoolbar
pyramid_tm
sqlalchemy.url = mysql://user:passwd#localhost/pyplay?charset=utf8
# By default, the toolbar only appears for clients from IP addresses
# '127.0.0.1' and '::1'.
debugtoolbar.hosts = 127.0.0.1 ::1
pish.theparam = somevalue
###
# wsgi server configuration
###
[server:main]
use = egg:waitress#main
host = 0.0.0.0
port = 6543
###
# logging configuration
# http://docs.pylonsproject.org/projects/pyramid/en/latest/narr/logging.html
###
[loggers]
keys = root, pyplay, sqlalchemy
[handlers]
keys = console
[formatters]
keys = generic
[logger_root]
level = INFO
handlers = console
[logger_pyplay]
level = DEBUG
handlers =
qualname = pyplay
[logger_sqlalchemy]
level = INFO
handlers =
qualname = sqlalchemy.engine
# "level = INFO" logs SQL queries.
# "level = DEBUG" logs SQL queries and results.
# "level = WARN" logs neither. (Recommended for production systems.)
[handler_console]
class = StreamHandler
args = (sys.stderr,)
level = NOTSET
formatter = generic
[formatter_generic]
format = %(asctime)s %(levelname)-5.5s [%(name)s][%(threadName)s] %(message)s
The reason it's difficult to do in pyramid is because it's always a bad idea to have module-level settings. It means your module can only ever be used in one way per-process (different code-paths can't use your library in different ways). :-)
A hack around not having access to the request object is to at least hide your global behind a function call, so that the global can be different per-thread (which is basically per-request).
def get_my_param(registry=None):
if registry is None:
registry = pyramid.threadlocals.get_current_registry()
return registry.settings['pyplay.theparam']
Step 1: create a singleton class say in a file xyz_file
class Singleton:
def __init__(self, klass):
self.klass = klass
self.instance = None
def __call__(self, *args, **kwds):
if self.instance == None:
self.instance = self.klass(*args, **kwds)
return self.instance
#Singleton
class ApplicationSettings(object):
def __init__(self, app_settings=None):
if app_settings is not None :
self._settings = app_settings
def get_appsettings_object(self):
return self
def get_application_configuration(self):
return self._settings
Step 2: in "__ init__.py"
def main(global_config, **settings):
....
.......
app_settings = ApplicationSettings(settings)
Step 3: You should be able to access in any part of the code.
from xyz_file import ApplicationSettings
app_settings = ApplicationSettings().get_application_configuration()
Basically, if you don't have access to the request object, you're "off the rails" in Pyramid. To do things the Pyramid way, we make components and figure out where they belong in the Pyramid lifecycle, and they should always have direct access to either or both of the registry (the ZCA) and the request.
If what you're doing doesn't fit in the the request lifecycle, then it's probably something that should be instantiated at server start up time, normally in your init.py where you build and fill the configurator (our access to the registry). Don't be afraid to use the registry to allow other components to get at things 'pseudo-globally' later. So probably you want to make some kind of factory for your thing, call the factory in your start up code, perhaps passing in a reference to the registry as an argument, and then attach the object to the registry. If your component needs to interface with request-lifecycle code, give it a method that takes request as a param. Later anything that needs at this object can get it from registry, and anything this object needs to get at can be done either through registry or request.
You can totally use the hack in the other answer to get at the current global registry, but needing to do so is a code smell, you can def figure out a better design to eliminate that.
pseudo code example, in the server start up code:
# in in the init block where our Configurator has been built
from myfactory import MyFactory
registry.my_component = MyFactory(config.registry)
# you can now get at my_component from anywhere in a pyramid system
your component:
class MyFactory(oject):
def __init__(self, registry):
# server start up lifecycle stuff here
self.registry = registry
def get_something(self, request):
# do stuff with the rest of the system
setting_wanted = self.registry.settings['my_setting']
Pyramid Views work this way. They are actually ZCA multi-adapters of request and context. Their factory is registered in the registry, and then when the view look up process kicks off, the factory instantiates a view passing in request as a param.
I have a complex Flask-based web app. There are lots of separate files with view functions. Their URLs are defined with the #app.route('/...') decorator. Is there a way to get a list of all the routes that have been declared throughout my app? Perhaps there is some method I can call on the app object?
All the routes for an application are stored on app.url_map which is an instance of werkzeug.routing.Map. You can iterate over the Rule instances by using the iter_rules method:
from flask import Flask, url_for
app = Flask(__name__)
def has_no_empty_params(rule):
defaults = rule.defaults if rule.defaults is not None else ()
arguments = rule.arguments if rule.arguments is not None else ()
return len(defaults) >= len(arguments)
#app.route("/site-map")
def site_map():
links = []
for rule in app.url_map.iter_rules():
# Filter out rules we can't navigate to in a browser
# and rules that require parameters
if "GET" in rule.methods and has_no_empty_params(rule):
url = url_for(rule.endpoint, **(rule.defaults or {}))
links.append((url, rule.endpoint))
# links is now a list of url, endpoint tuples
See Display links to new webpages created for a bit more information.
I just met the same question. Those solutions above are too complex.
Just open a new shell under your project:
>>> from app import app
>>> app.url_map
The first 'app' is my project script: app.py,
another is my web's name.
(this solution is for the tiny web with a little route)
I make a helper method on my manage.py:
#manager.command
def list_routes():
import urllib
output = []
for rule in app.url_map.iter_rules():
options = {}
for arg in rule.arguments:
options[arg] = "[{0}]".format(arg)
methods = ','.join(rule.methods)
url = url_for(rule.endpoint, **options)
line = urllib.unquote("{:50s} {:20s} {}".format(rule.endpoint, methods, url))
output.append(line)
for line in sorted(output):
print line
It solves the the missing argument by building a dummy set of options. The output looks like:
CampaignView:edit HEAD,OPTIONS,GET /account/[account_id]/campaigns/[campaign_id]/edit
CampaignView:get HEAD,OPTIONS,GET /account/[account_id]/campaign/[campaign_id]
CampaignView:new HEAD,OPTIONS,GET /account/[account_id]/new
Then to run it:
python manage.py list_routes
For more on manage.py checkout: http://flask-script.readthedocs.org/en/latest/
Apparently, since version 0.11, Flask has a built-in CLI. One of the built-in commands lists the routes:
FLASK_APP='my_project.app' flask routes
Similar to Jonathan's answer I opted to do this instead. I don't see the point of using url_for as it will break if your arguments are not string e.g. float
#manager.command
def list_routes():
import urllib
output = []
for rule in app.url_map.iter_rules():
methods = ','.join(rule.methods)
line = urllib.unquote("{:50s} {:20s} {}".format(rule.endpoint, methods, rule))
output.append(line)
for line in sorted(output):
print(line)
Use cli command in Directory where your flask project is.
flask routes
Since you did not specify that it has to be run command-line, the following could easily be returned in json for a dashboard or other non-command-line interface. The result and the output really shouldn't be commingled from a design perspective anyhow. It's bad program design, even if it is a tiny program. The result below could then be used in a web application, command-line, or anything else that ingests json.
You also didn't specify that you needed to know the python function associated with each route, so this more precisely answers your original question.
I use below to add the output to a monitoring dashboard myself. If you want the available route methods (GET, POST, PUT, etc.), you would need to combine it with other answers above.
Rule's repr() takes care of converting the required arguments in the route.
def list_routes():
routes = []
for rule in app.url_map.iter_rules():
routes.append('%s' % rule)
return routes
The same thing using a list comprehension:
def list_routes():
return ['%s' % rule for rule in app.url_map.iter_rules()]
Sample output:
{
"routes": [
"/endpoint1",
"/nested/service/endpoint2",
"/favicon.ico",
"/static/<path:filename>"
]
}
If you need to access the view functions themselves, then instead of app.url_map, use app.view_functions.
Example script:
from flask import Flask
app = Flask(__name__)
#app.route('/foo/bar')
def route1():
pass
#app.route('/qux/baz')
def route2():
pass
for name, func in app.view_functions.items():
print(name)
print(func)
print()
Output from running the script above:
static
<bound method _PackageBoundObject.send_static_file of <Flask '__main__'>>
route1
<function route1 at 0x128f1b9d8>
route2
<function route2 at 0x128f1ba60>
(Note the inclusion of the "static" route, which is created automatically by Flask.)
You can view all the Routes via flask shell by running the following commands after exporting or setting FLASK_APP environment variable.
flask shell
app.url_map
inside your flask app do:
flask shell
>>> app.url_map
Map([<Rule '/' (OPTIONS, HEAD, GET) -> helloworld>,
<Rule '/static/<filename>' (OPTIONS, HEAD, GET) -> static>])
print(app.url_map)
That, is, if your Flask application name is 'app'.
It's an attribute of the instance of the Flask App.
See https://flask.palletsprojects.com/en/2.1.x/api/#flask.Flask.url_map