getting 404 error using flask-restless

Question

I have an existing mysql database, and I'm trying to create an API to access it, but I'm receiving a 404 error. I have two versions of code that don't work.

Could someone please point me in the right direction?

URL I'm using: http://127.0.0.1:5000/api/email

v1:

from flask import Flask
from flask.ext.restless import APIManager
from flask_sqlalchemy import SQLAlchemy

app = Flask(__name__)
app.config['SQLALCHEMY_DATABASE_URI'] = 'mysql+pymysql://pathtodb/emails'
db = SQLAlchemy(app)

class Email(db.Model):
    __tablename__ = 'emails'
    id = db.Column(db.Integer, primary_key=True)
    status = db.Column(db.String(20))
    email = db.Column(db.String())

    def __init__(self, status, email):
        self.status = status
        self.email = email

api_manager = APIManager(app, flask_sqlalchemy_db=db)
api_manager.create_api(Email, methods=['GET', 'POST', 'DELETE', 'PUT'])

if __name__ == "__main__":
    app.run(debug=True)

v2:

from flask import Flask
from flask.ext.restless import APIManager
from sqlalchemy import *
from sqlalchemy.ext.declarative import declarative_base

app = Flask(__name__)
Base = declarative_base()
engine = create_engine('mysql+pymysql://pathtodb/emails', echo=True)
metadata = MetaData(bind=engine)

class Email(Base):
    __tablename__ = Table('emails', metadata, autoload=True)
    id = Column(Integer, primary_key=True)
    status = Column(String(20))
    email = Column(String())


api_manager = APIManager(app, flask_sqlalchemy_db=engine)
api_manager.create_api(Email, methods=['GET', 'POST', 'DELETE', 'PUT'])

if __name__ == "__main__":
    app.run(debug=True)

URL I'm using: http://127.0.0.1:5000/api/email

try with http://127.0.0.1:5000/api/emails, the tablename is plural — PRMoureu, Oct 31 '17 at 19:31
@PRMoureu I though I needed to use the class name 'email', but I need to use the table name as you suggested. I'd like to mark this as the answer, but I can't since it's a comment. Thank you, and I'll mark it as solved if you'd like to answer it. — SO03112, Oct 31 '17 at 19:51
i thought it was a typo, but you're right it could be confusing — PRMoureu, Oct 31 '17 at 19:57

score 4 · Accepted Answer · answered Oct 31 '17 at 19:57

4

With flask-restless the links are generated automatically when the method api_manager.create_api() is called, with this format :

'http://127.0.0.1:5000/api/{tablename}'  # default domain and port

where {tablename} is the __tablename__ given to the model class, not the class name called in the method.

The doc mentioned an example here.

answered Oct 31 '17 at 19:57

PRMoureu

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btw, not sure about the version you are using, but the import could be `from flask_restless import APIManager` to avoid a deprecated warning – PRMoureu Oct 31 '17 at 20:03

getting 404 error using flask-restless

1 Answers1