柯萊特單表繼承

我試圖端口SQLAlchemy的單表繼承例如柯萊特柯萊特單表繼承

class Person(Entity): 
    __tablename__ = 'people' 
    id = Column(sqlalchemy.Integer, primary_key=True) 
    discriminator = Column('type', sqlalchemy.String(50)) 
    __mapper_args__ = {'polymorphic_on': discriminator} 

    class Admin(EntityAdmin): 
     verbose_name = 'Person' 
     list_display = ['id','discriminator'] 

class Engineer(Person): 
    __mapper_args__ = {'polymorphic_identity': 'engineer'} 
    primary_language = Column(sqlalchemy.String(50)) 

    class Admin(EntityAdmin): 
     verbose_name = 'HR' 
     list_display = ['id','discriminator']

但是，這提供了以下錯誤：

Traceback (most recent call last): 
    File "D:/Programming/test/main.py", line 41, in <module> 
     start_application() 
    File "D:/Programming/test/main.py", line 37, in start_application 
     from testing.application_admin import MyApplicationAdmin 
    File "D:\Programming\test\testing\application_admin.py", line 6, in <module> 
     from testing.model import Engineer, Person 
    File "D:\Programming\test\testing\model.py", line 17, in <module> 
     class Engineer(Person): 
    File "C:\Python27\lib\site-packages\camelot\core\orm.py", line 334, in __init__ 
     super(EntityMeta, cls).__init__(classname, bases, dict_) 
    File "C:\Python27\lib\site-packages\sqlalchemy\ext\declarative.py", line 1343, in __init__ 
     _as_declarative(cls, classname, cls.__dict__) 
    File "C:\Python27\lib\site-packages\sqlalchemy\ext\declarative.py", line 1336, in _as_declarative 
     **mapper_args) 
    File "C:\Python27\lib\site-packages\sqlalchemy\orm\__init__.py", line 1129, in mapper 
     return Mapper(class_, local_table, *args, **params) 
    File "C:\Python27\lib\site-packages\sqlalchemy\orm\mapper.py", line 197, in __init__ 
     self._configure_inheritance() 
    File "C:\Python27\lib\site-packages\sqlalchemy\orm\mapper.py", line 473, in _configure_inheritance 
     self.local_table) 
    File "C:\Python27\lib\site-packages\sqlalchemy\sql\util.py", line 303, in join_condition 
     "between '%s' and '%s'.%s" % (a.description, b.description, hint)) 
    sqlalchemy.exc.ArgumentError: Can't find any foreign key relationships between 'people' and 'engineer'.

我做的東西是不是柯萊特方式？什麼是實現模型繼承的最佳方式？我無法在Camelot的文檔中找到任何內容。

編輯：

我試着下面添加到工程師類

__tablename__ = 'people' 
__table_args__ = {'extend_existing': True}

但現在經過我儘量選擇人或工程師表，或者如果我嘗試添加任何人，我得到：

C:\Python27\lib\site-packages\sqlalchemy\sql\expression.py:2276: SAWarning: Column u'people_id' on table <sqlalchemy.sql.expression.Select at 0x343fb90; Select object> being replaced by another column with the same key. Consider use_labels for select() statements. 
self[column.key] = column

任何提示？

來源

2013-04-08 rolisz

在工程師類

，該__tablename__需要被設置爲無，否則柯萊特將指定一個默認的表名和SQLAlchemy的將承擔多表繼承，因此需要一個外鍵關係。

來源

2013-04-12 19:05:24 Erik

柯萊特單表繼承

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