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
老问题，但由于这是谷歌中“sqlalchemy row to dict”的第一个结果，它值得一个更好的答案。
SqlAlchemy返回的RowProxy对象具有items()方法:
http://docs.sqlalchemy.org/en/latest/core/connections.html#sqlalchemy.engine.RowProxy.items
它只是返回一个(key, value)元组列表。因此可以使用以下方法将行转换为dict:
在Python中<= 2.6:
rows = conn.execute(query)
list_of_dicts = [dict((key, value) for key, value in row.items()) for row in rows]
在Python中>= 2.7:
rows = conn.execute(query)
list_of_dicts = [{key: value for (key, value) in row.items()} for row in rows]
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)
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}
返回this:class:的内容。KeyedTuple作为字典
In [46]: result = aggregate_events[0]
In [47]: type(result)
Out[47]: sqlalchemy.util._collections.result
In [48]: def to_dict(query_result=None):
...: cover_dict = {key: getattr(query_result, key) for key in query_result.keys()}
...: return cover_dict
...:
...:
In [49]: to_dict(result)
Out[49]:
{'calculate_avg': None,
'calculate_max': None,
'calculate_min': None,
'calculate_sum': None,
'dataPointIntID': 6,
'data_avg': 10.0,
'data_max': 10.0,
'data_min': 10.0,
'data_sum': 60.0,
'deviceID': u'asas',
'productID': u'U7qUDa',
'tenantID': u'CvdQcYzUM'}
