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我目前正在使用 Flask、SQLAlchemy 和 Marshmallow 构建一个 API。我有一个相当简单的数据库模型，如下所示：

class Transaction(db.Model):
    #Transaction Model
    __tablename__ = 'transactions'
    id = db.Column(db.Integer, primary_key = True)
    created_at = db.Column(db.DateTime, default=datetime.datetime.now)
    date = db.Column(db.DateTime)
    description = db.Column(db.Text)
    amount = db.Column(db.Float)
    category_id = db.Column(db.Integer, db.ForeignKey('categories.id'))
    category = db.relationship('Category', uselist=False, lazy='select')
    sub_category_id = db.Column(db.Integer, db.ForeignKey('sub_categories.id'))
    sub_category = db.relationship('SubCategory', uselist=False, lazy='select')
    account = db.Column(db.Integer, db.ForeignKey('accounts.id'))
    inner_transfer = db.Column(db.Boolean)



class Category(db.Model):
    __tablename__ = 'categories'
    id = db.Column(db.Integer, primary_key = True)
    category = db.Column(db.String(64), unique = True)
    transaction = db.relationship('Transaction')

我的 Marshmallow Schema 如下所示：

class CategorySchema(ma.ModelSchema):
    class Meta:
        model = Category


category_schema = CategorySchema()
categories_schema = CategorySchema(many=True)

class TransactionSchema(ma.ModelSchema):
    class Meta:
        model = Transaction
        include_fk = True

还有一个非常简单的 API 端点：

@api.route('/transactions', methods=['GET'])
def transactions():
    all_transactions = Transaction.query.all()
    for transaction in all_transactions:
        transaction.category = Category.query.filter_by(id=transaction.category_id).first()
        print(transaction.category)
    items, errors = transactions_schema.dump(all_transactions)
    return jsonify(items)

我现在的问题：如何构造一个包含类别名称而不是主键的 JSON 响应？我尝试了“所有事务中的事务”位，在打印语句中我得到了类别名称，但它没有显示在我的 JSON 响应中......

我不完全确定如何编写数据库模型。我特别坚持的是 db.relationship。在我的 Transaction 类中声明它似乎没有帮助，因为它也只显示 Category 类的主键。我上下阅读了 Flask-SQLAlchemy 文档，但我一无所知。

非常感谢任何帮助和指点！

谢谢！

4

1 回答 1

7

我创建了一个简单版本的模型来展示这一点，并在数据库中植入了两条测试记录：

from flask import Flask
from flask_sqlalchemy import SQLAlchemy
from flask_marshmallow import Marshmallow

app = Flask(__name__)
app.config['SQLALCHEMY_DATABASE_URI'] = 'sqlite:////tmp/test.db'
db = SQLAlchemy(app)
ma = Marshmallow(app)

# DB Models
class Transaction(db.Model):
    #Transaction Model
    __tablename__ = 'transactions'
    id = db.Column(db.Integer, primary_key = True)
    category_id = db.Column(db.Integer, db.ForeignKey('categories.id'))
    category = db.relationship('Category', uselist=False, lazy='select')

class Category(db.Model):
    __tablename__ = 'categories'
    id = db.Column(db.Integer, primary_key = True)
    category = db.Column(db.String(64), unique = True)
    transaction = db.relationship('Transaction')

# Schema
class TransactionSchema(ma.ModelSchema):
    class Meta:
        model = Transaction
        include_fk = True
    # Here is the function, you do not not need to define all fields you want to dump
    category_name = ma.Function(lambda obj: obj.category.category)

然后调用： TransactionSchema(many=True).dump(Transaction.query.all()).data

产量： [{u'category': 1, u'category_id': 1, u'category_name': u'Test1', u'id': 1}, {u'category': 2, u'category_id': 2, u'category_name': u'Test2', u'id': 2}]

也可以使用该功能覆盖您的字段并使其仅转储：

class TransactionSchema(ma.ModelSchema):
class Meta:
    model = Transaction
    include_fk = True
category = ma.Function(lambda obj: obj.category.category, dump_only=True)

产量： [{u'category': u'Test1', u'category_id': 1, u'id': 1}, {u'category': u'Test2', u'category_id': 2, u'id': 2}]

于 2016-09-20T15:48:16.617 回答