是否有一种简单的方法来迭代列名和值对？
我的sqlalchemy版本是0.5.6
下面是我尝试使用dict(row)的示例代码,但它抛出异常,TypeError:'User'对象不可迭代
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 …当我使用session.query时,我能够将结果转换为dicts列表:
my_query = session.query(table1,table2).filter(all_filters)
result_dict = [u.__dict__ for u in my_query.all()]
但是现在我必须使用该SELECT()操作,如何将结果转换为看起来像每个行结果的dict:
[{'Row1column1Name' : 'Row1olumn1Value', 'Row1column2Name' :'Row1Column2Value'},{'Row2column1Name' : 'Row2olumn1Value', 'Row2column2Name' : 'Row2Column2Value'},etc....].
这是我的SELECT()代码:
select = select([table1,table2]).where(all_filters)
res = conn.execute(select)
row = res.fetchone() #I have to use fetchone() because the query returns lots of rows
resultset=[]
while row is not None:
row = res.fetchone()
resultset.append(row)
print resultset
结果是:
[('value1', 'value2', 'value3', 'value4'),(.....),etc for each row]
我是Python的新手,任何帮助都会受到赞赏.