I am trying to expose a data service as an API for a PHP application. I have written the API logic in a file called APILogic.py. The code in this looks like this
class APILogic(object):
def __init__(self):
# the initialization is done here
...
def APIfunction(self):
# the API logic is built here, it uses the class variables and methods
...
I have created another file for the API purpose. Its called API.py. The code in this file looks like this
import APILogic from APILogic
class MyFlask:
def __init__(self):
self.test = APILogic()
from flask import Flask
app = Flask(__name__)
my_flask = MyFlask()
#app.route("/Test")
def Test():
return my_flask.test.APIfunction
if __name__ == "__main__":
app.run(debug=True,port=9999)
When I run the code, I get the error
> TypeError: APIfunction() takes 1 positional argument but 3 were given
The view function did not return a valid response. The return type must be a string, dict, tuple, Response instance, or WSGI callable, but it was a method.
There are no arguments for the APIfunction though.
Please help.
The view function did not return a valid response. The return type must be a string, dict, tuple, Response instance, or WSGI callable, but it was a method.
Looks like you're returning the method, but it sounds like you want to return the result of that method:
#app.route("/Test")
def Test():
return my_flask.test.APIfunction()
View function should return valid response.
Sample API code
from flask import Flask
app = Flask(__name__)
#app.route("/")
def hello():
return "Hello World!"
if __name__ == "__main__":
app.run()
Related
I have a simple code which returns render_template("home.html") of Flask in the main route. I wonder how can I test it using unittest?
#app.route("/", methods=["GET", "POST"])
def home():
return render_template("home.html")
I personally do the following
import unittest
from whereyourappisdefined import application
class TestFoo(unittest.TestCase):
# executed prior to each test
def setUp(self):
# you can change your application configuration
application.config['TESTING'] = True
# you can recover a "test cient" of your defined application
self.app = application.test_client()
# then in your test method you can use self.app.[get, post, etc.] to make the request
def test_home(self):
url_path = '/'
response = self.app.get(url_path)
self.assertEqual(response.status_code, 200)
For more information about testing Flask applications: https://flask.palletsprojects.com/en/2.0.x/testing/
I am building a Restful API in Python 3.8 using flask_restful, flask_apispec, and marshmallow. When I enable use_kwargs decorator my app is crushing on a POST with TypeError: post() takes 1 positional argument but 2 were given
I'd appreciate any help in solving this bug.
# requirements.txt
aniso8601==9.0.1
apispec==5.1.1
certifi==2021.10.8
charset-normalizer==2.0.7
click==8.0.3
Flask==2.0.2
flask-apispec==0.11.0
Flask-RESTful==0.3.9
gunicorn==20.1.0
idna==3.3
itsdangerous==2.0.1
Jinja2==3.0.2
MarkupSafe==2.0.1
marshmallow==3.14.0
pytz==2021.3
requests==2.26.0
six==1.16.0
tableauserverclient==0.17.0
urllib3==1.26.7
webargs==8.0.1
Werkzeug==2.0.2
from apispec import APISpec
from flask import Flask, request
from flask_restful import Resource, Api
from apispec.ext.marshmallow import MarshmallowPlugin
from flask_apispec.extension import FlaskApiSpec
from marshmallow import Schema, fields, post_load, ValidationError
from flask_apispec.views import MethodResource
from flask_apispec import use_kwargs, marshal_with
app = Flask(__name__) # Flask app instance initiated
api = Api(app) # Flask restful wraps Flask app around it.
app.config.update({
'APISPEC_SPEC': APISpec(
title='Kube Controller',
version='v1',
plugins=[MarshmallowPlugin()],
openapi_version='2.0.0'
),
'APISPEC_SWAGGER_URL': '/swagger/', # URI to access API Doc JSON
'APISPEC_SWAGGER_UI_URL': '/swagger-ui/' # URI to access UI of API Doc
})
docs = FlaskApiSpec(app)
class AwesomeRequestSchema(Schema):
api_type = fields.String(required=True)
#post_load
def create(self, data, **kwargs):
return MyAPI(**data)
class MyAPI:
def __init__(self, api_type):
self.api_type = api_type
self.message = "hi"
class AwesomeAPI(MethodResource, Resource):
#use_kwargs(AwesomeRequestSchema)
#marshal_with(AwesomeRequestSchema, code=200, description='Something created')
def post(self):
"""
POST
"""
try:
schema = AwesomeRequestSchema()
data = schema.load(request.json)
print(data.api_type)
return request.json
except ValidationError as err:
return err.messages
api.add_resource(AwesomeAPI, '/')
docs.register(AwesomeAPI)
if __name__ == '__main__':
app.run(debug=True)
Thanks!
I had the same issue. The use_kwargs decorator will try to populate and inject the AwesomeRequestSchema object into the post() method: https://flask-apispec.readthedocs.io/en/latest/api_reference.html#flask_apispec.annotations.use_kwargs
To fix, replace
def post(self):
with
def post(self, populated_request_object):
I want to build an api which accepts a parameter from the user which is a filepath and then process the file given in that path. The file to be processed is already in the server where the api will be running.
As of now, I have written an api where I have hardcoded the filepath in my code which runs the api. Now, I want to configure my api in such a way that accepts a filepath from the user. My api should accept the path as a parameter and process the file that has been given in the path.
The api code is as follows:
The convert function returns the category of the file.
import ectd
from ectd import convert
from flask import Flask, request
from flask_restful import Resource, Api
#from flask.views import MethodView
app = Flask(__name__)
api = Api(app)
#convert(r'D:\files\67cecf40-71cf-4fc4-82e1-696ca41a9fba.pdf')
class ectdtext(Resource):
def get(self, result):
return {'data': ectd.convert(result)}
#api.add_resource(ectdtext, '/ectd/<result>')
categories=convert(r'D:\files\6628cb99-a400-4821-8b13-aa4744bd1286.pdf')
#app.route('/')
def returnResult():
return categories
if __name__ == '__main__':
app.run(host="0.0.0.0", port=5000)
So, I want to make changes to this code to accept a parameter from the user which will be a filepath and the convert function will process that filepath. I want to know how to make my api accept a filepath parameter from the user.
Trial with requests.args.get:
import ectd
from ectd import convert
from flask import Flask, request
from flask_restful import Resource, Api
#from flask.views import MethodView
app = Flask(__name__)
api = Api(app)
#convert(r'D:\files\67cecf40-71cf-4fc4-82e1-696ca41a9fba.pdf')
class ectdtext(Resource):
def get(self, result):
return {'data': ectd.convert(result)}
#api.add_resource(ectdtext, '/ectd/<result>')
#app.route('/')
def returnResult():
categories=convert(r'D:\files\'.format(request.args.get('categories')))
return categories
if __name__ == '__main__':
app.run(host="0.0.0.0", port=5000)
results in error :
"RuntimeError: Working outside of request context.
This typically means that you attempted to use functionality that needed
an active HTTP request. Consult the documentation on testing for
information about how to avoid this problem."
PRESENT SCENARIO:
I am able to post a filepath to the url. My question is now how do I use this posted url with filepath in my code to trigger my function that takes in the filepath and processes the file. Code to post the filepath:
import ectd
from ectd import convert
from flask import Flask, request
from flask_restful import Resource, Api
#from flask.views import MethodView
app = Flask(__name__)
api = Api(app)
class ectdtext(Resource):
def get(self, result):
return {'data': ectd.convert(result)}
#api.add_resource(ectdtext, '/ectd/<result>')
categories=convert('/home/brian/ajay/files/5ca21af9-5b67-45f8-969c-ae571431c665.pdf')
#app.route('/')
def returnResult():
return categories
#app.route('/', defaults={'path': ''})
#app.route('/<path:path>')
def get_dir(path):
return path
##app.route('/get_dir/<path>')
#def get_dir(path):
# return path
if __name__ == '__main__':
app.run(host="0.0.0.0", port=5000)
This question already has an answer here:
Flask view raises TypeError: 'bool' object is not callable
(1 answer)
Closed 5 years ago.
I am trying to run a web service that takes ressource and account as arguments and returns the user pasword, but it returns me some errors :
raise ValueError('View function did not return a response')
ValueError: View function did not return a response
here is the stucture of my code :
from flask import Flask
import requests
# Config
TOKEN = ""
SERVER = ""
# Classes
class PClient():
def __init__(self, token=TOKEN, server=SERVER):
self.token = token
self.server = server
self.cache = {}
pass
def getPassword(self, resource, account):
if resource in self.cache and account in self.cache[resource]:
return self.cache[resource][account]
else :
return "Something"
# Script
app = Flask(__name__)
pmp = PClient()
#app.route('/<string:resource>/<string:account>')
def get_password(resource, account):
global pmp
return pmp.getPassword(resource, account)
if __name__ == "__main__":
app.run(host="127.0.0.1", port=2999, debug=True)
Any Idea?
Thank you
According to the Flask source, your view function returns None. You need to check your code that inserts values into self.cache[resource][account].
I'm learning Flask, and the request handling seems to be like:
#app.route("/")
def hello():
return "Hello World!"
So I end up defining the functions for all my routes in a single file. I'd much rather have functions for a model in its own file, e.g. get_user, create_user in user.py. I've used Express (node.js) in the past, and I can do:
user = require('./models/user')
app.get('/user', user.list)
where user.coffee (or .js) has a list function.
How do I do the same in Flask?
From the docs:
A decorator that is used to register a view function for a given URL rule. This does the same thing as add_url_rule() but is intended for decorator usage
The add_url_rule docs elaborate:
#app.route('/')
def index():
pass
Is equivalent to the following:
def index():
pass
app.add_url_rule('/', 'index', index)
You can just as easily import your view functions into a urls.py file and call add_url_rule once for each view function there instead of defining the rules along side the functions or use the lazy loading pattern.