Question

如何获得该关系另一端的类别的名称? 它是在建立关系时宣布的。我猜测,信息应当放在 s木图象中。

Let s say we have these three classes and a relations between them. Book * --- 1 Shelf and Book * --- * Author

class Shelf(Base):
    __tablename__ =  shelves 
    id = Column(Integer, primary_key = True)
    name = Column(String)
    #one-to-many books by backref in Book.shelf

    def __init__(self, name=""):
        self.name = name

class Book(Base):
    __tablename__ =  books 
    id = Column(Integer, primary_key = True)
    title = Column(String)
    #many-to-one shelf
    shelf_id = Column(Integer, ForeignKey( shelves.id ))
    shelf = relationship(Shelf, backref=backref( books , order_by=id))
    def __init__(self, title=""):
        self.title = title

author_book = Table( author_book , metadata, 
                    Column( author_id , Integer, ForeignKey( authors.id )),
                    Column( book_id , Integer, ForeignKey( books.id ))
                    )

class Author(Base):
    __tablename__ =  authors 
    id = Column(Integer, primary_key = True)
    name = Column(String)
    #many-to-many books
    books = relationship( Book , secondary=author_book, backref= authors )

    def __init__(self, name=""):
        self.name = name

From class_mapper(Class).iterate_properties we can easily get different properties: ColumnProperty and RelationshipProperty. There should be a way to get at leat the name of the class that is in relation. Any ideas?

Answer 1

也许没有什么次要的解决办法。

for prop in class_mapper(Shelf).iterate_properties:
    if isinstance(prop, sqlalchemy.orm.RelationshipProperty):
       print prop.mapper.class_

在两个方向与一对一对一对一对一对一对一对一对一对一。

Answer 2

0.8 0.8

from sqlalchemy import inspect
inspect(Shelf).relationships[ books ].mapper.class_

Answer 3

我找到了一种解决办法,但我并不相信,如果它始终发挥作用的话。在关系的另一端找到一个类别(类型)

class_mapper(Shelf)._props[ books ].mapper.class_

我们可能不了解属性名称(例如books in this case),但很难通过字典 * 寻找价值礼节,即sqlalchemy.orm.properties.Relationproperty。

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