[英]SQLAlchemy Partial Index with composite keys having the same name
I am using Python 2.7 and SQLAlchemy 0.9.8 and trying to use partial indexes. 我正在使用Python 2.7和SQLAlchemy 0.9.8,并尝试使用部分索引。
I have two tables (just the relevant part) 我有两个表(只是相关部分)
class DesignerRange
__tablename__ = 'designer_ranges'
shortcode = Column(Unicode(12))
class Design
__tablename__ = 'designs'
shortcode = Column(Unicode(12))
designer_range_id = Column(
Integer,
ForeignKey('designer_ranges.id'),
nullable=False,
)
designer_range = relationship(
DesignerRange,
backref=backref(
'designs',
single_parent=True,
),
lazy='joined',
)
And I want to create a partial Index that when the shortcode of the Design is not null (it exists) it should be unique within the DesignerRange. 我想创建一个局部索引,当Design的短代码不为null(它存在)时,它在DesignerRange中应该是唯一的。
I am trying something like this 我正在尝试这样的事情
@declarative.declared_attr
def __table_args__(cls):
return (
Index('design_shortcode',
table('designer_ranges', column('shortcode')).c.shortcode,
cls.shortcode, unique=True,
postgresql_where=(cls.shortcode!=None)),
)
This is the resulting Alembic migration 这就是由此产生的Alembic迁移
op.create_index('design_shortcode', 'designs', ['shortcode', 'shortcode'], unique=True, postgresql_where=sa.text('designs.shortcode IS NOT NULL'))
But I am getting this warning 但是我得到这个警告
sqlalchemy/sql/base.py:508: SAWarning: Column 'shortcode' on table <sqlalchemy.sql.selectable.TableClause at 0x7f50a1768550; designer_ranges> being replaced by Column('shortcode', Unicode(length=12), table=<designs>), which has the same key. Consider use_labels for select() statements.
I've no idea where to apply use_labes in this case, it seems to exist only within a select clause. 在这种情况下,我不知道在哪里应用use_labes,它似乎仅存在于select子句中。 Thanks
谢谢
I've solved my problem by using 我已经通过使用解决了我的问题
@declarative.declared_attr
def __table_args__(cls):
return (
Index('design_shortcode_idx',
cls.designer_range_id,
cls.shortcode, unique=True,
postgresql_where=(cls.shortcode!=None)),
)
So, instead of trying to get the shortcode from the foreign key table I am just getting its unique id. 因此,与其尝试从外键表中获取简码,不如获取其唯一ID。
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