I am trying to write a new python rest api like below,
from flask import Flask
from com.app.Controller import Event1LogicClass
from com.app.Controller import Event2LogicClass
from flask import Flask
from flask_restful import Api
com.app:
app = Flask(__name__)
apiRest = Api(app)
apiRest.add_resource(Event1LogicClass, '/Count/<string:days>')
apiRest.add_resource(Event2LogicClass,'/Count/<string:days>/<string:givenDate>')
if __name__ == '__main__':
app.run(debug = True)
com.app.Controller :
class Event1LogicClass(Resource)
def get(self, days):
return "Event1LogicClass"
com.app.Controller :
class Event2LogicClass(Resource)
def get(self, days, givenDate):
return "Event2LogicClass"
When I select MainClass as main class and run then It gives
resource_func = self.output(resource.as_view(endpoint, *resource_class_args,
AttributeError: module 'com.app.Controller.Event1LogicClass' has no attribute 'as_view'
I want to add multiple rest call in MainClass and call the api functionality depending on class in apiRest.add_resource.
where am I doing wrong?
Related
from flask import Flask, request
from flask_restplus import Resource, Api
app = Flask(name)
api = Api(app)
todos = {}
#api.route('/string:todo_id')
class TodoSimple(Resource):
def get(self, todo_id):
return {todo_id: todos[todo_id]}
def put(self, todo_id):
todos[todo_id] = request.form['data']
return {todo_id: todos[todo_id]}
if name == 'main':
app.run(debug=True)
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):
My Python Flask App is unable to decode the devanagari text eg:"सिंगापूर" that I pass as urlparam from my postman. If I sen english text it handles it well.
My PostMan query :
GET http://localhost:5000/getSimilarWord/सिंगापूर
from flask import Flask
from flask_restful import Resource, Api
class DevnagriText(Resource):
def get(self, textInput):
print("parsed String is :",textInput)
return {'text': textInput}
api.add_resource(DevnagriText, '/getWord/<string:textInput>')
if __name__ == '__main__':
app.run(debug=True)
The output I get on console is
{
"text": "\u00818"
}
Instead of
{
"text": "सिंगापूर"
}
You need to prevent the response from forcing ASCII: app.config['JSON_AS_ASCII'] = False
So in your example:
from flask import jsonify, Flask
from flask_restful import Resource, Api
app = Flask(__name__)
app.config['JSON_AS_ASCII'] = False
api = Api(app)
class DevnagriText(Resource):
def get(self, textInput):
print("parsed String is :",textInput)
return jsonify(text=textInput)
api.add_resource(DevnagriText, '/getWord/<string:textInput>')
if __name__ == '__main__':
app.run(debug=True)
But, this doesn't really matter, it will all get interpreted the same way if it gets read into Python or JavaScript.
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)
Ello ello,
I found similar questions on the bug i'm facing, and tried the solutions offered but it didn't work for me.
I'm trying to separate out my models in a different directory and import them into the app.py
When I try to import the db into the python terminal, i'm getting the no application found.
app.py code
from flask import Flask
from flask_restful import Resource, Api
# from flask_sqlalchemy import SQLAlchemy
from routes import test, root, user
from models.todo import db
app = Flask(__name__)
api = Api(app)
app.config['SQLALCHEMY_DATABASE_URI'] = 'postgresql://username:pass123#localhost/db'
app.config['SECRET_KEY'] = 'thiskeyissecret'
# db.init_app(app)
with app.app_context():
api = Api(app)
db.init_app(app)
api.add_resource(root.HelloWorld, '/')
api.add_resource(test.Test, '/test')
api.add_resource(user.User, '/user')
if __name__ == '__main__':
app.run(debug=True)
models
from flask_sqlalchemy import SQLAlchemy
db = SQLAlchemy()
class Todo(db.Model):
__tablename__ = 'Todos'
id = db.Column('id', db.Integer, primary_key=True)
data = db.Column('data', db.Unicode)
def __init__(self, id, data):
self.id = id
self.data = data
def __repr__(self):
return '<Todo %>' % self.id
my file directory looks like
Main_app
Models
Todo.py
routes
some routes
app.py
Flask-SQLAlchemy needs an active application context.
Try:
with app.app_context():
print(Todo.query.count())
From the flask documentation:
Purpose of the Context
The Flask application object has attributes, such as config, that are
useful to access within views and CLI commands. However, importing the
app instance within the modules in your project is prone to circular
import issues. When using the app factory pattern or writing reusable
blueprints or extensions there won’t be an app instance to import at
all.
Flask solves this issue with the application context. Rather than
referring to an app directly, you use the the current_app proxy, which
points to the application handling the current activity.
Flask automatically pushes an application context when handling a
request. View functions, error handlers, and other functions that run
during a request will have access to current_app.
It is ok to have db initialised in app.py
from flask import Flask
from flask_restful import Api
from flask_sqlalchemy import SQLAlchemy
from routes import test, root, user
app = Flask(__name__)
api = Api(app)
app.config['SQLALCHEMY_DATABASE_URI'] = 'postgresql://username:pass123#localhost/db'
app.config['SECRET_KEY'] = 'thiskeyissecret'
db = SQLAlchemy(app)
api.add_resource(root.HelloWorld, '/')
api.add_resource(test.Test, '/test')
api.add_resource(user.User, '/user')
if __name__ == '__main__':
app.run(debug=True)
Then in your todo.py
from app import db
class Todo(db.Model):
__tablename__ = 'Todos'
id = db.Column('id', db.Integer, primary_key=True)
data = db.Column('data', db.Unicode)
def __init__(self, id, data):
self.id = id
self.data = data
def __repr__(self):
return '<Todo %>' % self.id
I get a same err
that err reason for just can operation db in viewfunc
def __init__(self, id, data):
self.id = id
self.data = data
try move that code operation to your viewfunc
In a nutshell, do something like this:
from yourapp import create_app
app = create_app()
app.app_context().push()