My co-worker implemented an API that only allows GET requests with an ID parameter (so I can GET /foo/5 but can't GET /foo/). If I try to access the API's endpoint without providing an ID parameter, it (correctly) throws an unimplemented exception.
I want to fix this endpoint to show its documentation when viewed, without an ID, over the web. However, I still want it to throw an exception when that endpoint is accessed programatically.
As I remember it, django-rest-framework is capable of distinguishing those two cases (via request headers), but I'm not sure how to define the endpoint such that it returns either documentation HTML or an exception as appropriate.
Can anyone help provide the pattern for this?
Based on the description, I would guess that the endpoint is a function based view, which is registered on a route where it listens for get requests WITH parameters. I would suggest to register another route where you will listen for get requests without parameters...
from rest_framework.decorators import api_view
from rest_framework import status
#api_view(['GET'])
def existing_get_item_api(request, item_id, *args, **kwargs):
# query and return the item here ...
pass
#api_view(['GET'])
def get_help(request, *args, **kwargs):
# compose the help
return Response(data=help, status = status.HTTP_200_OK)
# somewhere in urls.py
urlpatterns = [
url(r'api/items/(?P<item_id>[0-9]+)/', existing_get_item_api),
url(r'api/items/', get_help),
]
Let me know how is this working out for you.
We can user modelviewsets and routers for this implementation
viewsets.py
class AccountViewSet(viewsets.ModelViewSet):
"""
A simple ViewSet for viewing and editing accounts.
"""
http_method_names = ['GET']
queryset = Account.objects.all()
serializer_class = AccountSerializer
routers.py
from rest_framework import routers
router = routers.SimpleRouter()
router.register(r'accounts', AccountViewSet)
Related
I am writing a REST API GET endpoint that needs to both return a response and store records to either GCP Cloud SQL (MySQL), but I want the return to not be dependent on completion of the writing of the records. Basically, my code will look like:
def predict():
req = request.json.get("instances")
resp = make_response(req)
write_to_bq(req)
write_to_bq(resp)
return resp
Is there any easy way to do this with Cloud SQL Client Library or something?
Turns our flask has a functionality that does what I require:
#app.route("predict", method=["GET"]):
def predict():
# do some stuff with the request.json object
return jsonify(response)
#app.after_request
def after_request_func(response):
# do anything you want that relies on context of predict()
#response.call_on_close
def persist():
# this will happen after response is sent,
# so even if this function fails, the predict()
# will still get it's response out
write_to_db()
return response
One important thing is that a method tagged with after_request must take an argument and return something of type flask.Response. Also I think if method has call_on_close tag, you cannot access from context of main method, so you need to define anything you want to use from the main method inside the after_request tagged method but outside (above) the call_on_close method.
I have a route that was created via flask_restplus.Api, not flask.Flask.
from flask import Flask
from flask_restplus import Api, Resource
flask_app = Flask(__name__)
app = Api(app=flask_app)
name_space = app.namespace('', description="Описание маршрутизаций")
And I would like to send an optional parameter like this, in the docs. Here is my code to send optional parameter:
from flask_restplus import Resource
#name_space.route("/structure/", defaults={'path': None})
#name_space.route("/structure/<string:path>")
#name_space.doc(params={'path': {'description': 'Get structure of a given path'}})
class struct_level(Resource):
def get(self, path):
print(path) # shows None
return method_to_work()
I've tried to change slashes in the end and to change method to post, nothing helped. Whatever value I send to 'path' parameter it returns None. If I make 'path' parameter required it works well, however I need this parameter to be optional, so that I can pass empty value.
Please, help me with an advice: how to make 'path' parameter optional in a route that was created by Api. Thanks!
P.S: Also, it is needed to be one endpoint.
I realised from my coworkers, that it is impossible to make this request in one route(endpoint). So I just had to create two endpoints, and that was ok.
I'm working in Flask and I want to allow users to enter information into a form without logging in but be required to login if they submit the form. After logging in, it should be as though a user just submitted the form (they shouldn't have to re-enter any information).
To store their information, I've used sessions like this. It works well:
if request.method == "POST":
if "arg1" not in session.keys() and "arg2" not in session.keys():
session["arg1"] = request.form.get('arg1')
session["arg2"] = request.form.get('arg2')
However, I'm having trouble with the login required part. I know I can use #login_required on the whole route but I just want #login_required to apply if the request is a post method. I've tried simply adding #login_required after checking if the method is a post request but it doesn't work.
My login route looks like this:
#app.route("/login", methods = ["POST", "GET"])
def login():
#log user in
return redirect(request.args.get("next") or url_for('index'))
It seems as though I need two things.
1: To apply #login_required solely to a post request.
2: To have request.args.get("next") call a post request, not get request
How could I go about doing these two things and achieve my goal?
Thank you!
Break out your routes. 1 for GET and 1 for POST
#app.route("/login", methods = ["GET"])
def get_login():
return stuff
#app.route("/login", methods = ["POST"])
#login_required
def post_login():
return stuff
There are a couple patterns that could be used here but this one is the most straight forward.
I need to make a custom controller on Odoo for getting information from the particular task. And I can able to produce the result also. But now I'm facing an issue.
The client needs to retrieve the information with a particular field.
For example,
The client needs to retrieve the information with the tracking number and the data must be JSON format also. If the tracking number is 15556456356, the url should be www.customurl.com/dataset/15556456356
The route of that URL should be #http.route('/dataset/<string:tracking_number>', type='http or json', auth="user or public"), basically the method should be like this:
import json
from odoo import http
from odoo.http import Response, request
class tracking(http.Controller):
# if user must be authenticated use auth="user"
#http.route('/dataset/<string:tracking_number>', type='http', auth="public")
def tracking(self, tracking_number): # use the same variable name
result = # compute the result with the given tracking_number and the result should be a dict to pass it json.dumps
return Response(json.dumps(result), content_type='application/json;charset=utf-8',status=200)
This method accept http request and return a json response, if the client is sending a json requests you should change type='json'. don't forget to import the file in the __init___.py.
Lets take an example let say that I want to return some information about a sale.order by a giving ID in the URL:
import json
from odoo import http
from odoo.http import Response, request
class Tracking(http.Controller):
#http.route('/dataset/<int:sale_id>', type='http', auth="public")
def tracking(self, sale_id):
# get the information using the SUPER USER
result = request.env['sale.order'].sudo().browse([sale_id]).read(['name', 'date_order'])
return Response(json.dumps(result), content_type='application/json;charset=utf-8',status=200)
So when I enter this URL using my Browser: http://localhost:8069/dataset/1:
I am using django-rest-framework's genericAPIViews
I want to send some data from the front end to the backend and depending upon the data sent Django should query a model and return some data to the frontend. The data sent is protected data and thus can't be attached in the URL so, GET request can't be used. I am not manipulating the database, just querying it and returning a response (a typical GET use case).
Now in DRF's genericAPIViews, I can't find a view which does this:
As can be seen from Tom Christie's GitHub page only 2 views have a post handler:
CreateAPIView: return self.create()
ListCreateAPIView: return self.create()
As can be seen both these views have post methods which create entries in the database which I don't want. Is there a built-in class which does my job or should I use generics.GenericAPIView and write my own post handler?
Currently I am using generic.View which has post(self, request, *args, **kwargs)
I think you have a few options to choose from. One way is to use a ModelViewSet which could be quite useful because of how it nicely handles the communication between views, serializers and models. Here is a link to django-rest-framework ModelViewSet docs.
These are the actions that it provides by default (since it inherits from GenericAPIView):
.list(), .retrieve(), .create(), .update(), .partial_update(), .destroy().
If you don't want all of them you could specify which methods you want by doing the following:
class ModelViewSet(views.ModelViewSet):
queryset = App.objects.all()
serializer_class = AppSerializer
http_method_names = ['get', 'post', 'head']
Note: http_method_names seems to be working from Django >= 1.8
Source: Disable a method in a ViewSet, django-rest-framework