Python SQLalchemy 找不到用于创建外键的表
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SQLalchemy not find table for creating foreign key
提问by Diego Moreira
I have a problem with SQL Alchemy, while trying to create a database, i get:
我在使用 SQL Alchemy 时遇到问题，在尝试创建数据库时，我得到：
"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'"
Meta datas:
元数据：
db = create_engine('postgresql+psycopg2:...//')
meta = MetaData()
meta.bind = db
Person table:
人表：
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 Table:
错误表：
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')
Normal table (creating normally the fk)
普通表（通常创建 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')
Creation Call:
创建调用：
meta.create_all(checkfirst=True)
Complete error log:
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'
采纳答案by Hamed
By adding the following line to my parenttable solved my problem. In case of Declarative:
通过将以下行添加到我的parent表中解决了我的问题。如果是声明式：
children = relationship("Child")
Otherwise: SQLAlchemy - Classic Mapper
Also try to have a look in here (SO)too, might help.
也尝试看看这里（SO），可能会有所帮助。
回答by Matthew Moisen
The solution is to replace the strings with actual columns:
解决方案是用实际列替换字符串：
Column('person_id', Integer, ForeignKey(tbl_person.c.id), primary_key=True)
回答by Jakobovski
In case of Declarative, I solved this problem by simply importing the class that was 'could not be found'.
在声明性的情况下，我通过简单地导入“无法找到”的类来解决这个问题。
