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
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'}
正如@balki提到的:
如果您正在查询特定的字段，可以使用_asdict()方法，因为它作为KeyedTuple返回。
In [1]: foo = db.session.query(Topic.name).first()
In [2]: foo._asdict()
Out[2]: {'name': u'blah'}
然而，如果您没有指定列，则可以使用其他建议的方法之一——例如@charlax提供的方法。注意，此方法仅对2.7+有效。
In [1]: foo = db.session.query(Topic).first()
In [2]: {x.name: getattr(foo, x.name) for x in foo.__table__.columns}
Out[2]: {'name': u'blah'}
Elixir是这样做的。这个解决方案的价值在于，它允许递归地包括关系的字典表示。
def to_dict(self, deep={}, exclude=[]):
"""Generate a JSON-style nested dict/list structure from an object."""
col_prop_names = [p.key for p in self.mapper.iterate_properties \
if isinstance(p, ColumnProperty)]
data = dict([(name, getattr(self, name))
for name in col_prop_names if name not in exclude])
for rname, rdeep in deep.iteritems():
dbdata = getattr(self, rname)
#FIXME: use attribute names (ie coltoprop) instead of column names
fks = self.mapper.get_property(rname).remote_side
exclude = [c.name for c in fks]
if dbdata is None:
data[rname] = None
elif isinstance(dbdata, list):
data[rname] = [o.to_dict(rdeep, exclude) for o in dbdata]
else:
data[rname] = dbdata.to_dict(rdeep, exclude)
return data
正在迭代的表达式求值为模型对象列表，而不是行。下面是正确的用法:
for u in session.query(User).all():
print u.id, u.name
你真的需要把它们转换成字典吗?当然，有很多方法，但是你不需要SQLAlchemy的ORM部分:
result = session.execute(User.__table__.select())
for row in result:
print dict(row)
更新:看一下sqlalchemy.orm.attributes模块。它有一组处理对象状态的函数，这可能对您很有用，特别是instance_dict()。
