python - sqlalchemy.exc.ProgrammingError : (psycopg2. ProgrammingError) 无法适配类型 'property'

Question

我正在尝试查询 Version，我希望将对象返回到它们的 max_version_number 属性等于 version_number 的位置:

latest_versions = \
    dbSession.query(Version).filter(Version.max_version_number == Version.number_version)

这会导致错误:

Traceback (most recent call last):
...
...
    filter(Version.max_version_number == Version.version_number).\
  File "c:\python27\lib\site-packages\sqlalchemy\orm\query.py", line 2588, in all
    return list(self)
  File "c:\python27\lib\site-packages\sqlalchemy\orm\query.py", line 2736, in __iter__
    return self._execute_and_instances(context)
  File "c:\python27\lib\site-packages\sqlalchemy\orm\query.py", line 2751, in _execute_and_instances
    result = conn.execute(querycontext.statement, self._params)
  File "c:\python27\lib\site-packages\sqlalchemy\engine\base.py", line 914, in execute
    return meth(self, multiparams, params)
  File "c:\python27\lib\site-packages\sqlalchemy\sql\elements.py", line 323, in _execute_on_connection
    return connection._execute_clauseelement(self, multiparams, params)
  File "c:\python27\lib\site-packages\sqlalchemy\engine\base.py", line 1010, in _execute_clauseelement
    compiled_sql, distilled_params
  File "c:\python27\lib\site-packages\sqlalchemy\engine\base.py", line 1146, in _execute_context
    context)
  File "c:\python27\lib\site-packages\sqlalchemy\engine\base.py", line 1341, in _handle_dbapi_exception
    exc_info
  File "c:\python27\lib\site-packages\sqlalchemy\util\compat.py", line 200, in raise_from_cause
    reraise(type(exception), exception, tb=exc_tb, cause=cause)
  File "c:\python27\lib\site-packages\sqlalchemy\engine\base.py", line 1139, in _execute_context
    context)
  File "c:\python27\lib\site-packages\sqlalchemy\engine\default.py", line 450, in do_execute
    cursor.execute(statement, parameters)
sqlalchemy.exc.ProgrammingError: (psycopg2.ProgrammingError) can't adapt type 'property' [SQL: 'SELECT "Versions".id AS "Versions_id", "Links".id AS "Links_id", "Entities".id AS "Entities_id", "SimpleEntities".stalker_version AS "SimpleEntities_stalker_version", "SimpleEntities".id AS "SimpleEntities_id", "SimpleEntities".entity_type AS "SimpleEntities_entity_type", "SimpleEntities".name AS "SimpleEntities_name", "SimpleEntities".description AS "SimpleEntities_description", "SimpleEntities".created_by_id AS "SimpleEntities_created_by_id", "SimpleEntities".updated_by_id AS "SimpleEntities_updated_by_id", "SimpleEntities".date_created AS "SimpleEntities_date_created", "SimpleEntities".date_updated AS "SimpleEntities_date_updated", "SimpleEntities".type_id AS "SimpleEntities_type_id", "SimpleEntities".generic_text AS "SimpleEntities_generic_text", "SimpleEntities".thumbnail_id AS "SimpleEntities_thumbnail_id", "SimpleEntities".html_style AS "SimpleEntities_html_style", "SimpleEntities".html_class AS "SimpleEntities_html_class", "Links".original_filename AS "Links_original_filename", "Links".full_path AS "Links_full_path", "Versions".task_id AS "Versions_task_id", "Versions".take_name AS "Versions_take_name", "Versions".version_number AS "Versions_version_number", "Versions".is_published AS "Versions_is_published", "Versions".created_with AS "Versions_created_with", "Versions".parent_id AS "Versions_parent_id" \nFROM "SimpleEntities" JOIN "Entities" ON "SimpleEntities".id = "Entities".id JOIN "Links" ON "Entities".id = "Links".id JOIN "Versions" ON "Links".id = "Versions".id \nWHERE "Versions".version_number = %(version_number_1)s'] [parameters: {'version_number_1': <property object at 0x00000000060EF6D8>}]

我做错了什么？

Here's 版本模型。

这是我的实际查询，没有问题:

proj_id = 5
versions = Version.query.join(Task).filter(Task.project_id == proj_id).all()

...但是，这会返回所有 版本，我希望只获取从查询返回的版本号最高的版本，这也是我尝试对其进行过滤的原因.

Answer 1

问题是您传递的是 property要比较的描述符对象(Version.max_version_number)，psycopg2 不知道如何适应。相反，您应该以一种或另一种方式使用 SQL 表达式。一种常见的方法是使用 hybrid attributes .

获取项目中按任务分组的最大版本行 project_id你可以use DISTINCT ON combined with ORDER BY在 PostgreSQL 中:

versions = Version.query\
    .join(Task)\
    .distinct(Version.task_id)\
    .filter(Task.project_id == proj_id)\
    .order_by(Version.task_id, Version.version_number.desc())\
    .all()

这将只返回 1 Version每Task虽然，即使 Version 的多个实例共享相同的 Version.version_number ，另一方面，这看起来很奇怪。

另一种选择是使用子查询作为排序查找并加入:

from sqlalchemy import tuple_

max_per_task = dbSession.query(Version.task_id,
                               func.max(Version.version_number).label('max'))\
    .join(Task)\
    .filter(Task.project_id == proj_id)\
    .group_by(Version.task_id)\
    .subquery()

versions = Version.query\
    .join(max_per_task,
          tuple_(max_per_task.c.task_id, max_per_task.c.max) ==
          tuple_(Version.task_id, Version.version_number))\
    .all()