python - 从类访问 SQLalchemy

Question

我正在做一个项目，我需要创建一个类实例，该实例具有连接到数据库并从中获取数据的方法(我使用 SQLite 作为后端)。我有一些使用 flask-sqlalchemy 的经​​验，但是当涉及到纯 SQLAlchemy 时我迷路了。概念如下:用户创建 DataSet 的实例，并将路径作为 __init__ 参数传递给数据库。如果数据库已经存在，我只想连接到它并进行查询，如果不存在，我想使用模型创建一个新数据库。但我不明白该怎么做。

这是 DataSet 代码:

from os.path import normcase, split, join, isfile
from sqlalchemy import create_engine
from sqlalchemy.orm import sessionmaker
import errors
import trainset
import testset


class DataSet:
    def __init__(self, path_to_set, path_to_db, train_set=False, path_to_labels=None, label_dict=None,
                 custom_name=None):
        self.__path_to_set = path_to_set
        self.__label_dict = label_dict

        if custom_name is None:
            dbpath = join(path_to_db, 'train.db')
            if train_set is False:
                dbpath = join(path_to_db, 'test.db')
        else:
            dbpath = join(path_to_db, custom_name)
        if isfile(dbpath):
            self.__prepopulated = True
        else:
            self.__prepopulated = False
        self.__dbpath = dbpath

        if train_set is True and path_to_labels is None:
            raise errors.InsufficientData('labels', 'specified')
        if train_set is True and not isfile(path_to_labels):
            raise errors.InsufficientData('labels', 'found at specified path', path_to_labels)

    def prepopulate(self):
        engine = create_engine('sqlite:////' + self.__dbpath)
        self.__prepopulated = True

这是trainset代码:

from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy import Column, String, PickleType, Integer, MetaData

Base = declarative_base()
metadata = MetaData()


class TrainSet(Base):
    __tablename__ = 'train set'
    id = Column(Integer, primary_key=True)
    real_id = Column(String(60))
    path = Column(String(120))
    labels = Column(PickleType)
    features = Column(PickleType)

这是测试集代码:

from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy import Column, String, PickleType, Integer, MetaData

Base = declarative_base()
metadata = MetaData()


class TestSet(Base):
    __tablename__ = 'test set'
    id = Column(Integer, primary_key=True)
    real_id = Column(String(60))
    path = Column(String(120))
    features = Column(PickleType)

因此，如果用户在创建 DataSet 实例时传递了 train_set=True，我想使用 TrainSet 模型创建数据库, 否则创建一个 TestSet 数据库。我希望这发生在 prepopulate 方法中，但是，我不知道该怎么做 - 文档要求这样做:Base.metadata.create_all(engine)，但我不知道将这段代码放在哪里。

Answer 1

首先保存参数train_set:

class DataSet:
    def __init__(self, path_to_set, path_to_db, train_set=False, path_to_labels=None, label_dict=None,
                 custom_name=None):
        self._train_set = train_set
        # ...

然后，在 prepopulate 中使用它来创建合适的模型:

def prepopulate(self):
    engine = create_engine('sqlite:////' + self.__dbpath)
    if self._train_set:
        trainset.Base.create_all(engine)
    else:
        testset.Base.create_all(engine)
    self.__prepopulated = True

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还有一件事:不要在您的“私有(private)”变量前加上双下划线。请阅读PEP 8 -- Style Guide for Python Code供引用。