item = Table('Item', metadata, autoload=True, autoload_with=engine, encoding = 'cp1257')
class Item(object):
pass
from sqlalchemy.orm import mapper
mapper(Item, item)
我收到错误:
line 43, in <module>
mapper(Item, item)
File "C:\Python27\lib\site-packages\sqlalchemy\orm\__init__.py", line 890, in mapper
return Mapper(class_, local_table, *args, **params)
File "C:\Python27\lib\site-packages\sqlalchemy\orm\mapper.py", line 211, in __init__
self._configure_properties()
File "C:\Python27\lib\site-packages\sqlalchemy\orm\mapper.py", line 578, in _configure_properties
setparent=True)
File "C:\Python27\lib\site-packages\sqlalchemy\orm\mapper.py", line 618, in _configure_property
self._log("_configure_property(%s, %s)", key, prop.__class__.__name__)
File "C:\Python27\lib\site-packages\sqlalchemy\orm\mapper.py", line 877, in _log
(self.non_primary and "|non-primary" or "") + ") " +
File "C:\Python27\lib\site-packages\sqlalchemy\util.py", line 1510, in __get__
obj.__dict__[self.__name__] = result = self.fget(obj)
File "C:\Python27\lib\site-packages\sqlalchemy\sql\expression.py", line 3544, in description
return self.name.encode('ascii', 'backslashreplace')
UnicodeDecodeError: 'ascii' codec can't decode byte 0xeb in position 7: ordinal not in range(128)
我正在连接到MSSQL。表autoload似乎工作。我只是在尝试映射时遇到此错误。 谢谢大家的帮助!
答案 0 :(得分:3)
将表映射到类会在类上创建映射属性。默认情况下,这些属性具有相同的列名称。因为python 2.x只允许使用ascii标识符,所以如果你有非ascii列名则会失败。
我能想到的唯一解决方案是在将表映射到类时为标识符指定不同的名称。
以下示例就是这样做的。请注意,为了简单起见,我在代码上创建表,因此任何人都可以在没有现有表的情况下运行代码。但你可以用反射表做同样的事情。
#-*- coding:utf-8 -*-
import sqlalchemy as sa
import sqlalchemy.orm
engine = sa.create_engine('sqlite://', echo=True) # new memory-only database
metadata = sa.MetaData(bind=engine)
# create a table. This could be reflected from the database instead:
tb = sa.Table('foo', metadata,
sa.Column(u'id', sa.Integer, primary_key=True),
sa.Column(u'nomé', sa.Unicode(100)),
sa.Column(u'ãéìöû', sa.Unicode(100))
)
tb.create()
class Foo(object):
pass
# maps the table to the class, defining different property names
# for some columns:
sa.orm.mapper(Foo, tb, properties={
'nome': tb.c[u'nomé'],
'aeiou': tb.c[u'ãéìöû']
})
之后,您可以使用Foo.nome
引用nomé
列,Foo.aeiou
引用ãéìöû
列。
答案 1 :(得分:1)
我遇到了同样的问题,最后设法替换了表['column']。自动加载后的密钥,只需让所有的表类继承这个,然后在mapTo方法中修改列名替换或手动覆盖使用字典和columns_descriptor方法的所需名称。我不知道这不是正确的方法,但是搜索几个小时后,我得到的是最好的方法。
class SageProxy(object):
@classmethod
def ismapped(cls, table_name=None):
if mappings:
if table_name:
if mappings.has_key(table_name):
tmap=mappings[table_name]
if tmap.has_key('class'):
tclass=tmap['class']
if tclass is cls:
return True
else:
for m in mappings:
if cls is m['class']:
return True
return False
@classmethod
def mappingprops(cls):
#override this to pass properties to sqlalchemy mapper function
return None
@classmethod
def columns_descriptors(cls):
#override this to map columns to different class properties names
#return dictionary where key is the column name and value is the desired property name
return {}
@classmethod
def mapTo(cls, table_name, map_opts=None):
if not cls.ismapped(table_name):
tab_obj=Table(table_name,sage_md,autoload=True)
for c in tab_obj.c:
#clean field names
tab_obj.c[c.name].key=c.key.replace(u'%',u'Porcentaje').replace(u'ñ',u'ny').replace(u'Ñ',u'NY').replace(u'-',u'_')
for k,v in cls.columns_descriptors():
if tab_obj.c[k]:
tab_obj.c[k].key=v
mapper(cls, tab_obj, properties=cls.mappingprops())
mappings[table_name]={'table':tab_obj,'class':cls}
return cls
我希望它有用
答案 2 :(得分:1)
我发现我可以通过对我反映的类的简单添加来做到这一点:
metadata = MetaData(bind=engine, reflect=True)
sm = sessionmaker(bind=engine)
class tblOrders(Base):
__table__ = metadata.tables['tblOrders']
meter = __table__.c['Meter#']
meter
现已映射到基础Meter#
列,该列允许此代码工作:
currOrder = tblOrders()
currOrder.meter = '5'
如果没有映射,python会将其视为一个损坏的语句,因为Meter
后跟对象中不存在注释。