SQLalchemy找不到用于创建外键的表 [英] SQLalchemy not find table for creating foreign key

问题描述

我在尝试创建数据库时遇到SQL Alchemy问题，

I have a problem with SQL Alchemy, while trying to create a database, i get:

"sqlalchemy.exc.NoReferencedTableError: Foreign key associated with column 'estate_agent.person_id' could not find table 'person' with which to generate a foreign key to target column 'id'"

元数据：

db = create_engine('postgresql+psycopg2:...//')
meta = MetaData()
meta.bind = db

人员表：

tbl_person = Table(
   'person', meta,
   Column('id', Integer, Sequence('seq_person_id'), primary_key=True),
   Column('name', String(100), unique=True, nullable = False),
   Column('password', String(40), nullable = False),
   Column('person_type_id', Integer, ForeignKey("person_type.id"), nullable = False),
   Column('register_date', DateTime, default = datetime.now),
   Column('pendencies', String(200)),
   Column('active', Boolean, default = True),
   schema = 'public')

Bug表格：

tbl_estate_agent = Table(
   'estate_agent', meta,
   Column('person_id', Integer, ForeignKey("person.id"), primary_key = True),
   Column('prize_range_id', Integer, ForeignKey("prize_range.id"), nullable = False),
   schema = 'public')

普通表（通常创建fk）

Normal table (creating normally the fk)

tbl_person_agent = Table(
   'person_agent', meta,
   Column('person_id', Integer, ForeignKey("person.id"), primary_key = True),
   Column('prize_range_id', Integer, ForeignKey("prize_range.id"), nullable = False),
   schema = 'public')

创建调用：

meta.create_all(checkfirst=True)

完整的错误日志：

Traceback (most recent call last):
   File "database_client.py", line 159, in <module>
    meta.create_all(checkfirst=True)
   File "/usr/local/lib/python2.7/dist-packages/sqlalchemy/sql/schema.py", line 3404, in create_all
    tables=tables)
   File "/usr/local/lib/python2.7/dist-packages/sqlalchemy/engine/base.py", line 1616, in _run_visitor
    conn._run_visitor(visitorcallable, element, **kwargs)
   File "/usr/local/lib/python2.7/dist-packages/sqlalchemy/engine/base.py", line 1245, in _run_visitor
    **kwargs).traverse_single(element)
   File "/usr/local/lib/python2.7/dist-packages/sqlalchemy/sql/visitors.py", line 120, in traverse_single
    return meth(obj, **kw)
   File "/usr/local/lib/python2.7/dist-packages/sqlalchemy/sql/ddl.py", line 699, in visit_metadata
    collection = [t for t in sort_tables(tables)
   File "/usr/local/lib/python2.7/dist-packages/sqlalchemy/sql/ddl.py", line 862, in sort_tables
    {'foreign_key': visit_foreign_key})
   File "/usr/local/lib/python2.7/dist-packages/sqlalchemy/sql/visitors.py", line 256, in traverse
    return traverse_using(iterate(obj, opts), obj, visitors)
   File "/usr/local/lib/python2.7/dist-packages/sqlalchemy/sql/visitors.py", line 247, in traverse_using
    meth(target)
   File "/usr/local/lib/python2.7/dist-packages/sqlalchemy/sql/ddl.py", line 853, in visit_foreign_key
    parent_table = fkey.column.table   File "/usr/local/lib/python2.7/dist-packages/sqlalchemy/util/langhelpers.py", line 725, in __get__
    obj.__dict__[self.__name__] = result = self.fget(obj)
   File "/usr/local/lib/python2.7/dist-packages/sqlalchemy/sql/schema.py", line 1720, in column tablekey)
sqlalchemy.exc.NoReferencedTableError: Foreign key associated with column 'estate_agent.person_id' could not find table 'person' with which to generate a foreign key to target column 'id'

推荐答案

通过将以下行添加到我的 parent 表解决了我的问题米在声明式的情况下：

By adding the following line to my parent table solved my problem. In case of Declarative:

children = relationship("Child")

否则： SQLAlchemy-经典映射器

也可以尝试在此处（SO）也可能有帮助。

Also try to have a look in here (SO) too, might help.

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