在 SQLAlchemy 中查询混合属性 [英] Querying a hybrid property in SQLAlchemy

问题描述

我将文件路径存储为数据库中的相对路径，但我随后使用混合属性在映射时将其转换为绝对路径.当我使用此属性进行查询时，它会引发错误.这是模型:

I'm storing file paths as relative paths in the database, but I'm then using hybrid properties to turn in into an absolute path when its mapped. When I query using this property it throws an error. Here's the model:

class File(Base):
    __tablename__ = 'files'
    ...

    _f_path = Column(Unicode(30))

    ...

    @hybrid_property
    def f_path(self):
        env = shelve.open('environment')
        return os.path.join(env['project_dir'], self._f_path)

    @f_path.setter
    def f_path(self, _f_path):
        self._f_path = _f_path

当我运行这个查询时(其中 ref 是一个 unicode 字符串):

When I run this query (where ref is a unicode string):

session.query(File).filter_by(f_path=ref).first()

它给了我这个错误:

File "/Users/Ben/Dropbox/Giraffe/giraffe_server/giraffe/file_handlers/maya.py", line 135, in process_file
    rf = session.query(File).filter_by(f_path=str(ref)).first()
  File "build/bdist.macosx-10.7-intel/egg/sqlalchemy/orm/query.py", line 1211, in filter_by
    for key, value in kwargs.iteritems()]
  File "build/bdist.macosx-10.7-intel/egg/sqlalchemy/orm/util.py", line 597, in _entity_descriptor
    return getattr(entity, key)
  File "build/bdist.macosx-10.7-intel/egg/sqlalchemy/ext/hybrid.py", line 681, in __get__
    return self.expr(owner)
  File "/Users/Ben/Dropbox/Giraffe/giraffe_server/giraffe/model.py", line 133, in f_path
    print "\n\n\n[model.py:File@f_path hybrid_property] returning: ", os.path.join(env['project_dir'], self._f_path)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/posixpath.py", line 66, in join
    if b.startswith('/'):
  File "build/bdist.macosx-10.7-intel/egg/sqlalchemy/sql/expression.py", line 3426, in __nonzero__
    raise TypeError("Boolean value of this clause is not defined")
TypeError: Boolean value of this clause is not defined

推荐答案

您的混合属性必须返回一个 sql 表达式；你的没有，它返回一个 python 字符串.

Your hybrid property must return a sql expression; yours does not, it returns a python string instead.

为了解决这个问题，不要在 python 中进行路径连接，而是在 SQL 表达式中:

To resolve that for this case, don't do the path join in python but in a SQL expression instead:

return env['project_dir'] + os.path.sep + self._f_path

将解析为 self._f_path.__radd__(result_of_project_dir_plus_os_path_sep)，既可用于查询，也可用作返回值.

which will resolve to self._f_path.__radd__(result_of_project_dir_plus_os_path_sep), which can be used both in queries and as a return value.

这篇关于在 SQLAlchemy 中查询混合属性的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！