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我有一个现有的 mysql 数据库,我正在尝试创建一个 API 来访问它,但我收到了 404 错误。我有两个版本的代码不起作用。

有人可以指出我正确的方向吗?

我正在使用的网址: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)

我正在使用的网址:http: //127.0.0.1 :5000/api/email

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1 回答 1

4

flask-restless调用方法时会自动生成链接,api_manager.create_api()格式如下:

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

{tablename}模型类的给定在哪里__tablename__,而不是方法中调用的类名。

文档在这里提到了一个例子。

于 2017-10-31T19:57:29.780 回答