我有一个SqlAlchemy查询对象,它使用uuid.UUID
import uuid
import sqlalchemy
foreign_uuid = '822965bb-c67e-47ee-ad12-a3b060ef79ae'
qry = Query(MyModel).filter(MyOtherModel.uuid == uuid.UUID(foreign_uuid))现在,我想从SQLAlchemy获得原始postgresql:
qry.statement.compile(dialect=postgresql.dialect(), compile_kwargs={"literal_binds": True}))这会产生以下错误:NotImplementedError: Don't know how to literal-quote value UUID('822965bb-c67e-47ee-ad12-a3b060ef79ae')
这似乎是因为SqlAlchemy只知道如何处理基本类型。文件显示使用TypeDecorator,甚至为GUID提供一个:
from sqlalchemy.dialects.postgresql import UUID
import uuid
class GUID(TypeDecorator):
"""Platform-independent GUID type.
Uses PostgreSQL's UUID type, otherwise uses
CHAR(32), storing as stringified hex values.
"""
impl = CHAR
def load_dialect_impl(self, dialect):
if dialect.name == 'postgresql':
return dialect.type_descriptor(UUID())
else:
return dialect.type_descriptor(CHAR(32))
def process_bind_param(self, value, dialect):
if value is None:
return value
elif dialect.name == 'postgresql':
return str(value)
else:
if not isinstance(value, uuid.UUID):
return "%.32x" % uuid.UUID(value).int
else:
# hexstring
return "%.32x" % value.int
def process_result_value(self, value, dialect):
if value is None:
return value
else:
if not isinstance(value, uuid.UUID):
value = uuid.UUID(value)
return value然而,它并没有说明如何使用这个TypeDecorator。我需要在我的模型中引用它吗?我是否以某种方式使它可用于.statement.compile呼叫?
发布于 2021-05-26 16:35:00
使用修饰类型作为列类型。另一组文档更清楚地解释了它:https://docs.sqlalchemy.org/en/14/faq/sqlexpressions.html#faq-sql-expression-string
from sqlalchemy import TypeDecorator, Integer
class MyFancyType(TypeDecorator):
impl = Integer
def process_literal_param(self, value, dialect):
return "my_fancy_formatting(%s)" % value
from sqlalchemy import Table, Column, MetaData
tab = Table('mytable', MetaData(), Column('x', MyFancyType()))https://stackoverflow.com/questions/54284604
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