Question

通过Flask SQL Alchemy，我正在使用Chinook sqlite数据库。

sqlalchemy.exc.ArgumentError：映射器映射的类PlayLists-> playlists无法为映射表“ playlists”组合任何主键列

我的代码是这样的。 “ app / 初始化 .py”

from flask import Flask
from config import app_config
from flask_sqlalchemy import SQLAlchemy
from flask_bootstrap import Bootstrap
db = SQLAlchemy()

def create_app(config_name):
    app = Flask(__name__, instance_relative_config=True)
    app.config.from_object(app_config[config_name])
    app.config.from_pyfile('config.py')
    db.app = app
    db.init_app(app)
    db.Model.metadata.reflect(db.engine)
    Bootstrap(app)

    from app import models

    return app

app / model.py

from app import db


class PlayLists(db.Model):
    __tablename__ = db.Model.metadata.tables['playlists']

我在做什么错了？

Answer 1

在播放列表类中，您将db.Model.metadata.tables ['playlists']分配给 __ tablename __ 。但是，db.Model.metadata.tables ['playlists']返回类'sqlalchemy.sql.schema.Table'的对象。您应该改为将其分配给具有表名称的字符串，如：

app / model.py

from app import db


class PlayLists(db.Model):
    __tablename__ = 'playlists'

此示例对我有用，返回反射数据库的列名：

from flask import Flask
from flask_sqlalchemy import SQLAlchemy

app = Flask(__name__)
app.secret_key = 'SUPERSECRET'
app.config['SQLALCHEMY_DATABASE_URI'] = ''mysql+pymysql://user:pass@localhost:port/db''

db = SQLAlchemy(app)
db.init_app(app)
db.Model.metadata.reflect(db.engine)


class User(db.Model):
    __tablename__ = "users"


@app.route("/")
def hello():
    user = User()
    table_columns = str(user.__table__.columns)
    return table_columns


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

Flask SQL Alchemy反射无法组装任何主键

1 个答案: