Question

我有一个基于SQLAlchemy的现有项目，具有以下结构

# __init__.py

import yaml
from sqlalchemy.orm import sessionmaker
from sqlalchemy import create_engine
from sqlalchemy.ext.declarative import declarative_base


try:
    with open('config.yaml', 'r') as yaml_config_file:
        config = yaml.safe_load(yaml_config_file)
except:
    raise Exception

Base = declarative_base()
Session = sessionmaker(bind=create_engine(config['db']))


# models.py:

   from fmgcore import Base
   class Image(Base):
     __tablename_- = 'image'
     ....
     ...



# service.py

from fmgcore import config, Session
import sqlalchemy.exc as sql_exc

log = logging.getLogger(__name__)


class Services(object):
  def __init__(self, config):
    self.config = config

  def create_image(self, name, filename,
                 disk_format,
                 size=None,
                 description=None):

    session = Session()

    try:
        session.add(...)
        session.commit(...)

    except sql_exc.IntegrityError as e:
        logging.info("Image with name {0} already exists".format(name))
        session.rollback()
    except sql_exc.SQLAlchemyError as e:
        session.rollback()
    finally:
        session.close()


def get_image():
    ####

我想使用Django-Rest使用REST Api公开这些服务。我不想使用Django模型。

我的问题是我应该如何在view.py中使用Services（），以便它不会为每个REST请求创建引擎或新连接的多个实例，而是使用同一池。

将基于SQLAlchemy的项目与Django-rest集成

0 个答案: