Example of RESTful API in Flask Python - python

Can someone show me examples of making a RESTful API which uses database information in Flask? I have no idea how to implement POST, PUT and DELETE and I always get the 405 error where I can't use the method in url.

Have you add request method in your routing? you can following reference from: flask-restful
from flask import Flask, request
from flask_restful import Resource, Api
app = Flask(__name__)
api = Api(app)
class TodoSimple(Resource):
def get(self):
# do get something
def put(self):
# do put something
def delete(self):
# do delete something
def post(self):
# do post something
api.add_resource(TodoSimple, '/api/todo')
if __name__ == '__main__':
app.run(debug=True)

in flask-restful the HTTP actions (GET, PUT, POST, DELETE) have their corresponding method in the resource class, so is just a matter of defining those method in the resource (with the corresponding parameter defined in the routing)
I've also built a lightweight framework for building restful apis that makes it super easy to build apis. You can take a look at the code to have an idea of how an API can be built, configured and run, and of course, build on top of it
here's the code: https://github.com/sebastiandev/peach

Related

Define a common route pass-through or filter for all Python Flask Blueprint Routes

In my Python Flask app I have Blueprint Routes defined that correspond to API endpoints,
# Submit Agreement (Insert or Update)
#bp.route('submitAgreement', methods=['POST'])
#auth.login_required
def submitAgreement():
#...code...
# Get Existing Agreement
#bp.route('fetchAgreement', methods=['POST'])
#auth.login_required
def fetchAgreement():
#...code...
I need to define a common route pass-through or filter that will perform authorization prior to executing the code. The authorization needs to be, if the URL contains the param id=.., check that that ID belongs to the logged-in user.
Is there a way to define a custom "aspect" or filter with this code in Python Flask?
Flask offers an
#app.before_request decorator
You can substitute app for bp to apply the function you define to just the blueprint.

webapp2 + RESTful API

I am building on Google App Engine + Python + webapp2. Part of building a modern web app requires a restful API. I know that I can do this with Flask, however I want to explore the possibility of building a REST API on webapp2.
On webapp2, requests are handled like this:
app = webapp2.WSGIApplication([
('/post/new', CreatePost),
('/post/([a-z0-9]+)', ViewPost),
('/post/([a-z0-9]+)/edit', EditPost),
('/post/([a-z0-9]+)/delete', DeletePost)
])
Note: ([a-z0-9]+) is a regex that represents the post_id
The above request handlers do not adhere to a RESTful pattern since the request methods are specified in the path (/delete, /edit, /new) rather than in the request headers.
Is the solution to create a single handler class that receives all request types? For example:
class PostHandler(webapp2.RequestHandler):
def get(self):
# handle GET requests
def post(self):
# handle POST requests
def put(self):
# handle PUT requests
def delete(self):
# handle DELETE requests
app = webapp2.WSGIApplication([
('/post/?', PostHandler)
])
In this case, all /post paths are handled by PostHandler. The post_id is no longer used in this pattern since it would be submitted in the request body instead.
Is this the correct approach to building a REST API with webapp2?
You are on the right path, but you should continue to handle post_id in the url and do it like this:
class PostHandler(webapp2.RequestHandler):
def get(self, post_id=None):
if post_id:
# handle Fetching a single post object
else:
# handle Queries
def post(self, post_id=None):
if post_id:
self.abort(405)
# handle creating a single post object
def put(self, post_id=None):
if post_id:
# handle updating a single post object
else:
self.abort(405)
def delete(self, post_id=None):
if post_id:
# handle deleting a single post object
else:
self.abort(405)
app = webapp2.WSGIApplication([
('/post/<post_id>/', PostHandler),
('/post/', PostHandler),
])
Furthermore putting the HTTP verb within the request payload like voscausa suggested is not inline with RESTful API design.
One should extend webapp2.RequestHandler and use the extension as the new base class for a RESTfulHandler and then maybe also for your specific purpose extend the RESTfulHandler to a RESTfulPostHandler. This way we could also deal with new ways or work such as jwt (JSON web token) in the header, authorization and other properties that you wish to handle in the header.
BTW, "post" might not be the best name for your item/object since it could easily get mixed up with the http verb "post". If I were you, I would rename it into "item" or similar, and use a RESTfulItemHandler which extends the webapp2 request handler. There are many things you would like to share between your handler and like that it is convenient to share the functionality in a base-class.

Injecting a Flask Request into another Flask App

Is there a way to inject a Flask request object into a different Flask app. This is what I'm trying to do:
app = flask.Flask(__name__)
#app.route('/foo/<id>')
def do_something(id):
return _process_request(id)
def say_hello(request):
# request is an instance of flask.Request.
# I want to inject it into 'app'
I'm trying this with Google Cloud Functions, where say_hello() is a function that is invoked by the cloud runtime. It receives a flask.Request as the argument, which I want to then process through my own set of routes.
I tried the following, which doesn't work:
def say_hello(request):
with app.request_context(request.environ):
return app.full_dispatch_request()
This responds with 404 errors for all requests.
Edit:
The simple way to implement say_hello() is as follows:
def say_hello(request):
if request.method == 'GET' and request.path.startswith('/foo/'):
return do_something(_get_id(request.path))
flask.abort(404)
This essentially requires me to write the route matching logic myself. I'm wondering if there's a way to avoid doing that, and instead use Flask's built-in decorators and routing capabilities.
Edit 2:
Interestingly, dispatching across apps work locally:
app = flask.Flask(__name__)
# Add app.routes here
functions = flask.Flask('functions')
#functions.route('/', defaults={'path': ''})
#functions.route('/<path:path>', methods=['GET', 'POST', 'PUT', 'DELETE'])
def catch_all(path):
with app.request_context(flask.request.environ):
return app.full_dispatch_request()
if __name__ == '__main__':
functions.run()
But the same technique doesn't seem to work on GCF.
I wouldn't recommend this method, but this is technically possible by abusing the request stack and rewriting the current request and re-dispatching it.
However, you'll still need to do some type of custom "routing" to properly set the url_rule, as the incoming request from GCF won't have it (unless you explicitly provide it via the request):
from flask import Flask, _request_ctx_stack
from werkzeug.routing import Rule
app = Flask(__name__)
#app.route('/hi')
def hi(*args, **kwargs):
return 'Hi!'
def say_hello(request):
ctx = _request_ctx_stack.top
request = ctx.request
request.url_rule = Rule('/hi', endpoint='hi')
ctx.request = request
_request_ctx_stack.push(ctx)
return app.dispatch_request()

Flask-Restful ignores blueprint's url_prefix

I want to use Flask blueprints to organize my Flask-Restful resources into separate url prefixes. Regardless of what url prefix I set (during either blueprint creation or registration), everything gets mapped to the raw route paths. How do I correctly use Restful with blueprints?
app = Flask(__name__)
api = Api(app)
api.add_resource(Building, '/<int:id>', endpoint='building')
api.add_resource(Jack, '/<int:id>', endpoint='jack')
building_api = Blueprint('building_api', __name__)
jack_api = Blueprint('jack_api', __name__)
app.register_blueprint(building_api, url_prefix='/buildings')
app.register_blueprint(jack_api, url_prefix='/jacks')
All documentation I can find says that these should now be available at /buildings/<int:id> and /jacks/<int:id>, but both of those urls 404 and instead I can access the building one at /<int:id>. Hard coding the path in add_resource fixes it, but defeats the point of url_prefix.
You need to pass the blueprint to the Api instance, not the app.
building_bp = Blueprint('buildings', __name__)
building_api = Api(building_bp)
building_api.add_resource(Building, '/<int:id>')
app.register_blueprint(building_bp, url_prefix='/buildings')
This is zhe best way to do with blueprint:
from flask import Flask, Blueprint
from flask_restful import Api, Resource, url_for
app = Flask(__name__)
api_bp = Blueprint('api', __name__)
api = Api(api_bp)
class TodoItem(Resource):
def get(self, id):
return {'task': 'Say "Hello, World!"'}
api.add_resource(TodoItem, '/todos/<int:id>')
app.register_blueprint(api_bp)
you should send Blueprint'instance to Api
I do not know why but I struggled using the Blueprint as mentioned in the answers.
But here's a quick solution I found while going through the doc link. Making use of the prefix parameter in Api() does the job.
app = Flask(__name__)
api = Api(app, prefix='/buildings')
Now, all your routes will be prefixed with /buildings. Just make sure you use url_for('link') in places where you might have simply used a /link.
One more strange thing I noticed is that atleast for me, it did not work until I renamed my routes to the same name as their class names. For example, Class Home(Resource) should have a route to /home. Using /homeepage or any other route for Home Class causes an error. Not sure if it is only me.

What is an 'endpoint' in Flask?

The Flask documentation shows:
add_url_rule(*args, **kwargs)
Connects a URL rule. Works exactly like the route() decorator.
If a view_func is provided it will be registered with the endpoint.
endpoint – the endpoint for the registered URL rule. Flask itself assumes the name of the view function as endpoint
What exactly is meant by an "endpoint"?
How Flask Routing Works
The entire idea of Flask (and the underlying Werkzeug library) is to map URL paths to some logic that you will run (typically, the "view function"). Your basic view is defined like this:
#app.route('/greeting/<name>')
def give_greeting(name):
return 'Hello, {0}!'.format(name)
Note that the function you referred to (add_url_rule) achieves the same goal, just without using the decorator notation. Therefore, the following is the same:
# No "route" decorator here. We will add routing using a different method below.
def give_greeting(name):
return 'Hello, {0}!'.format(name)
app.add_url_rule('/greeting/<name>', 'give_greeting', give_greeting)
Let's say your website is located at 'www.example.org' and uses the above view. The user enters the following URL into their browser:
http://www.example.org/greeting/Mark
The job of Flask is to take this URL, figure out what the user wants to do, and pass it on to one of your many python functions for handling. It takes the path:
/greeting/Mark
...and matches it to the list of routes. In our case, we defined this path to go to the give_greeting function.
However, while this is the typical way that you might go about creating a view, it actually abstracts some extra info from you. Behind the scenes, Flask did not make the leap directly from URL to the view function that should handle this request. It does not simply say...
URL (http://www.example.org/greeting/Mark) should be handled by View Function (the function "give_greeting")
Actually, it there is another step, where it maps the URL to an endpoint:
URL (http://www.example.org/greeting/Mark) should be handled by Endpoint "give_greeting".
Requests to Endpoint "give_greeting" should be handled by View Function "give_greeting"
Basically, the "endpoint" is an identifier that is used in determining what logical unit of your code should handle the request. Normally, an endpoint is just the name of a view function. However, you can actually change the endpoint, as is done in the following example.
#app.route('/greeting/<name>', endpoint='say_hello')
def give_greeting(name):
return 'Hello, {0}!'.format(name)
Now, when Flask routes the request, the logic looks like this:
URL (http://www.example.org/greeting/Mark) should be handled by Endpoint "say_hello".
Endpoint "say_hello" should be handled by View Function "give_greeting"
How You Use the Endpoint
The endpoint is commonly used for the "reverse lookup". For example, in one view of your Flask application, you want to reference another view (perhaps when you are linking from one area of the site to another). Rather than hard-code the URL, you can use url_for(). Assume the following
#app.route('/')
def index():
print url_for('give_greeting', name='Mark') # This will print '/greeting/Mark'
#app.route('/greeting/<name>')
def give_greeting(name):
return 'Hello, {0}!'.format(name)
This is advantageous, as now we can change the URLs of our application without needing to change the line where we reference that resource.
Why not just always use the name of the view function?
One question that might come up is the following: "Why do we need this extra layer?" Why map a path to an endpoint, then an endpoint to a view function? Why not just skip that middle step?
The reason is because it is more powerful this way. For example, Flask Blueprints allow you to split your application into various parts. I might have all of my admin-side resources in a blueprint called "admin", and all of my user-level resources in an endpoint called "user".
Blueprints allow you to separate these into namespaces. For example...
main.py:
from flask import Flask, Blueprint
from admin import admin
from user import user
app = Flask(__name__)
app.register_blueprint(admin, url_prefix='admin')
app.register_blueprint(user, url_prefix='user')
admin.py:
admin = Blueprint('admin', __name__)
#admin.route('/greeting')
def greeting():
return 'Hello, administrative user!'
user.py:
user = Blueprint('user', __name__)
#user.route('/greeting')
def greeting():
return 'Hello, lowly normal user!'
Note that in both blueprints, the '/greeting' route is a function called "greeting". If I wanted to refer to the admin "greeting" function, I couldn't just say "greeting" because there is also a user "greeting" function. Endpoints allow for a sort of namespacing by having you specify the name of the blueprint as part of the endpoint. So, I could do the following...
print url_for('admin.greeting') # Prints '/admin/greeting'
print url_for('user.greeting') # Prints '/user/greeting'
Endpoint is the name used to reverse-lookup the url rules with url_for and it defaults to the name of the view function.
Small example:
from flask import Flask, url_for
app = Flask(__name__)
# We can use url_for('foo_view') for reverse-lookups in templates or view functions
#app.route('/foo')
def foo_view():
pass
# We now specify the custom endpoint named 'bufar'. url_for('bar_view') will fail!
#app.route('/bar', endpoint='bufar')
def bar_view():
pass
with app.test_request_context('/'):
print url_for('foo_view')
print url_for('bufar')
# url_for('bar_view') will raise werkzeug.routing.BuildError
print url_for('bar_view')
If you have same class name and want to map with multiple routes, then specify the endpoint, so that framework will differentiate between two:
class ClassName(Resource):
def get(self):
if request.endpoint!='hello':
return {"data": "Hello"}
elif:
return {"data" : "World"}
api.add_resource(ClassName, '/rout1', endpoint = "world")
api.add_resource(ClassName, '/rout2', endpoint="hello")
#app.route('/') #Endpoint
def a_function(): #View function
return 'view'
Inside Flask, every endpoint with its request methods mapped to a view function. When you use app.route decorator you are actually adding a URL rule.

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