在Python中与Singleton共享声明性_基(SQLAlchemy)
当所有内容都在一个文件中时,我可以正常运行SQLAlchemy。我现在想把我的模型放到另一个文件中 但是,这不起作用,因为我找不到一种分享基地的方式。我尝试使用单例,但model.py中的单例为Null,并且从未在数据库中创建模式 我该怎么做才能解决这个问题 我的文件(简化版): main/main.py:在Python中与Singleton共享声明性_基(SQLAlchemy),python,sql,singleton,sqlalchemy,Python,Sql,Singleton,Sqlalchemy,当所有内容都在一个文件中时,我可以正常运行SQLAlchemy。我现在想把我的模型放到另一个文件中 但是,这不起作用,因为我找不到一种分享基地的方式。我尝试使用单例,但model.py中的单例为Null,并且从未在数据库中创建模式 我该怎么做才能解决这个问题 我的文件(简化版): main/main.py: from model import User from utils.utils import readConf,createSession,getBase class Worker(thr
from model import User
from utils.utils import readConf,createSession,getBase
class Worker(threading.Thread):
def __init__(self, queue, session):
self.__queue = queue
threading.Thread.__init__(self)
self._stopevent = threading.Event( )
def run(self):
session.merge(User(queue.get()))
session.commit()
class Collector(threading.Thread):
def __init__(self, nom = ''):
threading.Thread.__init__(self)
self.nom = nom
self._stopevent = threading.Event( )
def run(self):
while not self._stopevent.isSet():
queue.put("Name")
if __name__ == '__main__':
conf = readConf("file.")
session = createSession(conf)
queue = Queue.Queue(0)
Worker(queue, session).start()
Collector("Start").start()
utils/utils.py
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.schema import MetaData
def createSession(conf):
schema = conf['bdd']['type'] + '://' + conf['bdd']['user'] + ':' + conf['bdd']['password'] + '@' + conf['bdd']['host'] + '/' + conf['bdd']['db']
engine = create_engine(schema, echo=True)
b = getBase("Utils")
b.set_base(declarative_base())
Base = b.get_base()
Base.metadata.create_all(engine)
Session = sessionmaker(bind=engine)
session = Session()
return session
class getBase(object):
__single = None # the one, true Singleton
__base = None
__text = None
__base = None
def __new__(cls, *args, **kwargs):
# Check to see if a __single exists already for this class
# Compare class types instead of just looking for None so
# that subclasses will create their own __single objects
if cls != type(cls.__single):
cls.__single = object.__new__(cls, *args, **kwargs)
__base = declarative_base()
return cls.__single
def __init__(self,name=None):
self.name = name
def get_base(self):
return self.__base
def set_base(self, value):
self.__base = value
model/model.py
from utils.utils import getBase
b = getBase("model")
b.set_base(declarative_base())
Base = b.get_base()
class User(Base):
__tablename__ = 'users'
def __init__(self, name):
self.screen_name = name
name_id = Column(Integer, primary_key=True, autoincrement=True)
screen_name = Column(BigInteger(), primary_key=True, nullable=False)
编辑@Lafaya
我更改了model/\uuuuu init\uuuuuu.py
如下:
#!/usr/bin/python
Base = declarative_base()
from model import Base
class User(Base):
etc...
然后我更改了model/model.py
如下:
#!/usr/bin/python
Base = declarative_base()
from model import Base
class User(Base):
etc...
现在我无法从model.py导入Base(因为它可以通过\uuu init.py\uu
导入)。但我需要一个参考基地
Traceback (most recent call last):
File "../main/main.py", line 12, in <module>
from model.model import User
File "../model/model.py", line 3, in <module>
from model import Base
ImportError: cannot import name Base
回溯(最近一次呼叫最后一次):
文件“./main/main.py”,第12行,在
从model.model导入用户
文件“./model/model.py”,第3行,在
从模型导入库
ImportError:无法导入名称基
在模型包的\uuuuu init\uuuuuuuuuupy
中写入基础。然后您可以导入并使用它
将Base
放入model/\uuuu init\uuuu.py
好的,我终于将模型放入init.py了