小编Jia*_*ang的帖子

鉴于我有以下客户端哈希,是否有一个快速的ruby方式(无需编写多行脚本)来获取密钥给定我想匹配client_id？例如,如何获得密钥client_id == "2180"？

clients = {
  "yellow"=>{"client_id"=>"2178"}, 
  "orange"=>{"client_id"=>"2180"}, 
  "red"=>{"client_id"=>"2179"}, 
  "blue"=>{"client_id"=>"2181"}
}

我正在使用SQLAlchemy的具体表继承.在声明样式模型类中,我已成功配置它.

我的代码就像:

class Entry(AbstractConcreteBase, db.Model):
    """Base Class of Entry."""

    id = db.Column(db.Integer, primary_key=True, nullable=False)
    created = db.Column(db.DateTime, nullable=False)
    post_id = declared_attr(lambda c: db.Column(db.ForeignKey("post.id")))
    post = declared_attr(lambda c: db.relationship("Post", lazy="joined"))

    @declared_attr
    def __tablename__(cls):
        return cls.__name__.lower()

    @declared_attr
    def __mapper_args__(cls):
        # configurate subclasses about concrete table inheritance
        return {'polymorphic_identity': cls.__name__,
                'concrete': True} if cls.__name__ != "Entry" else {}

class TextEntry(Entry):
    """Text and Article Entry."""

    text = db.deferred(db.Column(db.Text, nullable=False))

class PhotoEntry(Entry):
    """Photo Entry."""

    path = db.deferred(db.Column(db.String(256), nullable=False))

它在shell中测试时工作正常:

>>> from models.entry import Entry
>>> …