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}
在python 3.8+中，我们可以使用数据类和它附带的asdict方法来实现这一点:
from dataclasses import dataclass, asdict
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import sessionmaker
from sqlalchemy import Column, String, Integer, create_engine
Base = declarative_base()
engine = create_engine('sqlite:///:memory:', echo=False)
@dataclass
class User(Base):
__tablename__ = 'users'
id: int = Column(Integer, primary_key=True)
name: str = Column(String)
email = Column(String)
def __init__(self, name):
self.name = name
self.email = 'hello@example.com'
Base.metadata.create_all(engine)
SessionMaker = sessionmaker(bind=engine)
session = SessionMaker()
user1 = User("anurag")
session.add(user1)
session.commit()
query_result = session.query(User).one() # type: User
print(f'{query_result.id=:}, {query_result.name=:}, {query_result.email=:}')
# query_result.id=1, query_result.name=anurag, query_result.email=hello@example.com
query_result_dict = asdict(query_result)
print(query_result_dict)
# {'id': 1, 'name': 'anurag'}
关键是使用@dataclass装饰器，并用它的类型(name: str = column (String)行的:str部分)注释每一列。
还要注意，由于电子邮件没有注释，因此它不包括在query_result_dict中。
我是一个新晋的Python程序员，遇到了使用join表获取JSON的问题。使用这里的答案中的信息，我构建了一个函数，将合理的结果返回到JSON，其中包括表名，避免使用别名或字段冲突。
简单地传递会话查询的结果:
test = Session()。查询(VMInfo、客户). join(客户).order_by (VMInfo.vm_name) .limit (50) .offset (10)
json = sqlAl2json（test）
def sqlAl2json(self, result):
arr = []
for rs in result.all():
proc = []
try:
iterator = iter(rs)
except TypeError:
proc.append(rs)
else:
for t in rs:
proc.append(t)
dict = {}
for p in proc:
tname = type(p).__name__
for d in dir(p):
if d.startswith('_') | d.startswith('metadata'):
pass
else:
key = '%s_%s' %(tname, d)
dict[key] = getattr(p, d)
arr.append(dict)
return json.dumps(arr)
