[FIXED] Flask-restful AttributeError: type object 'Project' has no attribute 'as_view'

Issue

So im still learning about the structure in flask it seems flask give a too much flexibilty which is kinda confusing for me to decide, so im trying to add my Resource class to the API with the current structure

api
  /__init__.py
  /project.py
models
  /project_management.py
configs.py
configs.json
run.py

but it return error when running AttributeError: type object 'Project' has no attribute 'as_view'

Traceback (most recent call last):

File "e:\project-py\run.py", line 6, in from api import * File "e:\project-py\api_init_.py", line 9, in from .project import * File "e:\project-py\api\project.py", line 6, in @api.add_resource(Project, '/v1/project') File "C:\Users\rokie\Anaconda3\envs\py39-rest-flask\lib\site-packages\flask_restful_init_.py", line 391, in add_resource
self.register_view(self.app, resource, *urls, **kwargs) File "C:\Users\rokie\Anaconda3\envs\py39-rest-flask\lib\site-packages\flask_restful_init.py", line 431, in _register_view
resource_func = self.output(resource.as_view(endpoint, *resource_class_args, AttributeError: type object 'Project' has no attribute 'as_view'

My run.py

from functools import wraps
from flask import Flask, session, g, render_template, flash
# from flask_cors import CORS, cross_origin
from flask_wtf.csrf import CSRFProtect
from pages import *
from api import *

from config import create_app

app = create_app()
app.app_context().push()

# app.register_blueprint(pages)
app.register_blueprint(api, url_prefix='/api')

# app.secret_key = "ff82b98ef8727e388ea8bff063"
csrf = CSRFProtect()
csrf.init_app(app)


if __name__ == "__main__":
    app.run(host='127.0.0.1',debug=True)

Config.py

import json
from flask import Flask
from flask_sqlalchemy import SQLAlchemy
from flask_marshmallow import Marshmallow
# from flask_wtf.csrf import CSRFProtect
from flask_login import LoginManager
from flask_cors import CORS, cross_origin


db = SQLAlchemy()
ma = Marshmallow()

f = open('configs.json')
config = json.load(f)


def create_app():
    app = Flask(__name__, template_folder='templates')
    app.config['SQLALCHEMY_DATABASE_URI'] = config['config']['database']
    app.config['SECRET_KEY'] = config['config']['secret_key']

    app.config['SQLALCHEMY_TRACK_MODIFICATIONS'] = False

    api_v1_cors_config = {
        "origins": ["*"]
    }
    CORS(app, resources={
        r"/api/*": api_v1_cors_config
    })

    db.init_app(app)
    ma.init_app(app)

    login_manager = LoginManager()
    login_manager.login_view = 'pages.login'
    login_manager.init_app(app)

    from models.user_management import User

    @login_manager.user_loader
    def load_user(user_id):
        # since the user_id is just the primary key of our user table, use it in the query for the user
        return User.query.get(int(user_id))
    return app

api init.py folder with the name 'api' as the package

from flask import Blueprint
from flask_restful import Api

api_bp=Blueprint('api',__name__)
api = Api(api_bp)

# from .login import *
# from .api_project import *
from .project import *
# from .master_user import *

and lastly the project.py

from . import api
from flask import Flask, jsonify, request
from flask_restful import Resource, Api
from models.project_management import Project, ProjectStatus, Task

@api.add_resource(Project, '/v1/project')
class Project(Resource):
    def get(self):
        try:
            print('test')
            projects = Project.query.all()
            return jsonify({
                'message': 'Data get success',
                'data': projects,
            })
        except Exception as e:
            return jsonify({
                'message': f'failed to get data {e}',
                'data': []
            })

enter code here

Solution

I can't check it but ...

You have the same name Project for model (Project.query.all()) and for class class Project(Resource) and this can make problem.

Because model Project is created (imported) before add_resource() so it uses this model but it has to use class Project

You may have to first define class with different name and later use add_resource() using new class

class MyProject(Resource):
   # ... code ...  

api.add_resource(MyProject, '/v1/project')

Answered By - furas