在Flask Admin中创建Flask-SQLAlchemy的Brand模型时触发AttributeError

在Flask-SQLAlchemy中定义了Product和Brand两个关联模型，尝试在Flask Admin中创建Brand时，触发以下错误，无法完成创建：

AttributeError: 'tuple' object has no attribute 'items'

相关代码如下：

from flask import Flask
from flask_sqlalchemy import SQLAlchemy
from datetime import datetime
from flask_admin import Admin
from flask_admin.contrib.sqla import ModelView


app = Flask(__name__)
app.config['SQLALCHEMY_DATABASE_URI'] = 'postgresql+psycopg2://postgres:postgres@localhost:5432/ecommerce'
app.config['SQLALCHEMY_TRACK_MODIFICATIONS'] = False
app.config['SECRET_KEY'] = 'secret'
db = SQLAlchemy(app)



# Models
class Brand(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(100), nullable=False, unique=True)
    products = db.relationship('Product', backref='brand', lazy=True)

    def __str__(self):
        return self.name

def format_products(view, context, model, name):
    return ', '.join([product.name for product in model.products])

class BrandView(ModelView):
    column_list = ['name', 'products']
    column_formatters = {'products': format_products}
    form_excluded_columns = ['products']

-

class Product(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(200), nullable=False)
    description = db.Column(db.Text)
    price = db.Column(db.Float, nullable=False)
    brand_id = db.Column(db.Integer, db.ForeignKey('brand.id'), nullable=False)



with app.app_context():
    db.create_all()

admin = Admin(app)
admin.add_view(ModelView(Product, db.session))
admin.add_view(BrandView(Brand, db.session))


@app.route('/')
def hello_world():
    return 'Hello World!'


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

错误原因 #

代码中BrandView类定义结束后，存在一个多余的-符号，这个符号破坏了Python代码的结构解析逻辑，导致后续Product类的定义被错误解析，最终在Flask Admin初始化模型视图时触发了元组相关的属性错误。

修复方案 #

移除BrandView类和Product类之间的多余-符号，确保代码结构合法。

修复后的完整代码：

from flask import Flask
from flask_sqlalchemy import SQLAlchemy
from datetime import datetime
from flask_admin import Admin
from flask_admin.contrib.sqla import ModelView


app = Flask(__name__)
app.config['SQLALCHEMY_DATABASE_URI'] = 'postgresql+psycopg2://postgres:postgres@localhost:5432/ecommerce'
app.config['SQLALCHEMY_TRACK_MODIFICATIONS'] = False
app.config['SECRET_KEY'] = 'secret'
db = SQLAlchemy(app)



# Models
class Brand(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(100), nullable=False, unique=True)
    products = db.relationship('Product', backref='brand', lazy=True)

    def __str__(self):
        return self.name

def format_products(view, context, model, name):
    return ', '.join([product.name for product in model.products])

class BrandView(ModelView):
    column_list = ['name', 'products']
    column_formatters = {'products': format_products}
    form_excluded_columns = ['products']

class Product(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(200), nullable=False)
    description = db.Column(db.Text)
    price = db.Column(db.Float, nullable=False)
    brand_id = db.Column(db.Integer, db.ForeignKey('brand.id'), nullable=False)



with app.app_context():
    db.create_all()

admin = Admin(app)
admin.add_view(ModelView(Product, db.session))
admin.add_view(BrandView(Brand, db.session))


@app.route('/')
def hello_world():
    return 'Hello World!'


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

内容的提问来源于stack exchange，提问作者Kashif Raza Khan