是否有一种简单的方法来遍历列名和值对?
我的SQLAlchemy版本是0.5.6
下面是我尝试使用dict(row)的示例代码:
import sqlalchemy
from sqlalchemy import *
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import sessionmaker
print "sqlalchemy version:",sqlalchemy.__version__
engine = create_engine('sqlite:///:memory:', echo=False)
metadata = MetaData()
users_table = Table('users', metadata,
Column('id', Integer, primary_key=True),
Column('name', String),
)
metadata.create_all(engine)
class User(declarative_base()):
__tablename__ = 'users'
id = Column(Integer, primary_key=True)
name = Column(String)
def __init__(self, name):
self.name = name
Session = sessionmaker(bind=engine)
session = Session()
user1 = User("anurag")
session.add(user1)
session.commit()
# uncommenting next line throws exception 'TypeError: 'User' object is not iterable'
#print dict(user1)
# this one also throws 'TypeError: 'User' object is not iterable'
for u in session.query(User).all():
print dict(u)
在我的系统输出上运行这段代码:
Traceback (most recent call last):
File "untitled-1.py", line 37, in <module>
print dict(u)
TypeError: 'User' object is not iterable
为了完成@Anurag Uniyal的回答，这里有一个递归地遵循关系的方法:
from sqlalchemy.inspection import inspect
def to_dict(obj, with_relationships=True):
d = {}
for column in obj.__table__.columns:
if with_relationships and len(column.foreign_keys) > 0:
# Skip foreign keys
continue
d[column.name] = getattr(obj, column.name)
if with_relationships:
for relationship in inspect(type(obj)).relationships:
val = getattr(obj, relationship.key)
d[relationship.key] = to_dict(val) if val else None
return d
class User(Base):
__tablename__ = 'users'
id = Column(Integer, primary_key=True)
first_name = Column(TEXT)
address_id = Column(Integer, ForeignKey('addresses.id')
address = relationship('Address')
class Address(Base):
__tablename__ = 'addresses'
id = Column(Integer, primary_key=True)
city = Column(TEXT)
user = User(first_name='Nathan', address=Address(city='Lyon'))
# Add and commit user to session to create ids
to_dict(user)
# {'id': 1, 'first_name': 'Nathan', 'address': {'city': 'Lyon'}}
to_dict(user, with_relationship=False)
# {'id': 1, 'first_name': 'Nathan', 'address_id': 1}
您可以像这样将sqlalchemy对象转换为字典，并将其作为json/dictionary返回。
辅助功能:
import json
from collections import OrderedDict
def asdict(self):
result = OrderedDict()
for key in self.__mapper__.c.keys():
if getattr(self, key) is not None:
result[key] = str(getattr(self, key))
else:
result[key] = getattr(self, key)
return result
def to_array(all_vendors):
v = [ ven.asdict() for ven in all_vendors ]
return json.dumps(v)
驱动程序功能:
def all_products():
all_products = Products.query.all()
return to_array(all_products)
我对马可·马里亚尼(Marco Mariani)的回答有一个变体，以装饰者的身份表达。主要的区别是它将处理实体列表，以及安全地忽略一些其他类型的返回值(这在使用mock编写测试时非常有用):
@decorator
def to_dict(f, *args, **kwargs):
result = f(*args, **kwargs)
if is_iterable(result) and not is_dict(result):
return map(asdict, result)
return asdict(result)
def asdict(obj):
return dict((col.name, getattr(obj, col.name))
for col in class_mapper(obj.__class__).mapped_table.c)
def is_dict(obj):
return isinstance(obj, dict)
def is_iterable(obj):
return True if getattr(obj, '__iter__', False) else False
你在你的项目中到处都需要它，我很欣赏@anurag的回答，它很好。直到这一点上，我正在使用它，但它会混乱你所有的代码，也不会与实体改变工作。
不如试试这个，
继承SQLAlchemy中的基查询类
from flask_sqlalchemy import SQLAlchemy, BaseQuery
class Query(BaseQuery):
def as_dict(self):
context = self._compile_context()
context.statement.use_labels = False
columns = [column.name for column in context.statement.columns]
return list(map(lambda row: dict(zip(columns, row)), self.all()))
db = SQLAlchemy(query_class=Query)
在那之后，无论你在哪里定义你的对象“as_dict”方法都会在那里。
