[英]SQLAlchemy: order by a relationship field in a relationship
在我正在研究的金字塔應用程序中,我有以下場景:
class Widget(Base):
__tablename__ = 'widgets'
id = Column(Integer, primary_key=True)
name = Column(String(50))
sidebar = Column(mysql.TINYINT(2))
def __init__(self, name, sidebar):
self.name = name
self.sidebar = sidebar
class Dashboard(Base):
__tablename__ = 'dashboard'
user_id = Column(Integer, ForeignKey('users.id'), primary_key=True)
widget_id = Column(Integer, ForeignKey('widgets.id'), primary_key=True)
delta = Column(mysql.TINYINT)
widget = relationship('Widget')
def __init__(self, user_id, widget_id, delta):
self.user_id = user_id
self.widget_id = widget_id
self.delta = delta
class User(Base):
__tablename__ = 'users'
id = Column(Integer, primary_key=True)
login = Column(Unicode(255), unique=True)
password = Column(Unicode(60))
fullname = Column(Unicode(100))
dashboard = relationship('Dashboard', order_by='Dashboard.widget.sidebar, Dashboard.delta')
def __init__(self, login, password, fullname):
self.login = login
self.password = crypt.encode(password)
self.fullname = fullname
因此,我希望用戶的“儀表板”關系具有用戶的儀表板記錄,但是由“側欄”(這是儀表板的關系屬性)排序。 目前我收到此錯誤:
sqlalchemy.exc.InvalidRequestError: Property 'widget' is not an instance of ColumnProperty (i.e. does not correspond directly to a Column).
在關系聲明中這種排序是否可行?
謝謝!
有了這個,試着想一想SQL SQLAlchemy在嘗試加載User.dashboard時應該發出什么。 就像SELECT * FROM dashboard JOIN widget ... ORDER BY widget.sidebar
? 或SELECT * FROM dashboard ORDER BY (SELECT sidebar FROM widget...
?通過不同的表對結果進行排序,對於relationship()
的工作來說過於開放,以決定它自己。這可以通過提供的方式來實現在條件列表達式的Dashboard
,可以提供這種排序,當ORM發出簡單的SELECT對儀表板的表,以及當它是指它在一個不那么簡單,因為選擇在那里它可能跨用戶,儀表盤表加盟立刻(例如急切加載)。
我們提供自定義SQL表達式,特別是那些涉及其他表的表達式,使用column_property() ,或者當我們不希望默認情況下加載表達式時( 或許在這種情況下可能) ,使用deferred( )。 例:
from sqlalchemy import *
from sqlalchemy.orm import *
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()
class Widget(Base):
__tablename__ = 'widgets'
id = Column(Integer, primary_key=True)
name = Column(String(50))
sidebar = Column(Integer)
class Dashboard(Base):
__tablename__ = 'dashboard'
user_id = Column(Integer, ForeignKey('users.id'), primary_key=True)
widget_id = Column(Integer, ForeignKey('widgets.id'), primary_key=True)
delta = Column(Integer)
widget = relationship('Widget')
widget_sidebar = deferred(select([Widget.sidebar]).where(Widget.id == widget_id))
class User(Base):
__tablename__ = 'users'
id = Column(Integer, primary_key=True)
login = Column(Unicode(255), unique=True)
dashboard = relationship('Dashboard', order_by='Dashboard.widget_sidebar, Dashboard.delta')
e = create_engine("sqlite://", echo=True)
Base.metadata.create_all(e)
s = Session(e)
w1, w2 = Widget(name='w1', sidebar=1), Widget(name='w2', sidebar=2)
s.add_all([
User(login='u1', dashboard=[
Dashboard(
delta=1, widget=w1
),
Dashboard(
delta=2, widget=w2
)
]),
])
s.commit()
print s.query(User).first().dashboard
“.dashboard”加載發出的最終SQL是:
SELECT dashboard.user_id AS dashboard_user_id, dashboard.widget_id AS dashboard_widget_id, dashboard.delta AS dashboard_delta
FROM dashboard
WHERE ? = dashboard.user_id ORDER BY (SELECT widgets.sidebar
FROM widgets
WHERE widgets.id = dashboard.widget_id), dashboard.delta
請記住,MySQL在為上述子查詢進行優化方面做得非常糟糕 。 如果您需要高性能,可以考慮將“側欄”的值復制到“儀表板”中,即使這會使一致性更難以維護。
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