【发布时间】:2013-05-06 17:19:37
【问题描述】:
我正在通过创建新项目来尝试 Pyramid。我选择 PostgreSQL 和 sqlalchemy。现在我有一个手动创建的表格“照片”和一个模型:
class Photo(Base):
""" The SQLAlchemy declarative model class for a Photo object. """
__tablename__ = 'photo'
id = Column(Integer, primary_key=True)
name = Column(Text)
filename = Column(Text)
cat_id = Column(Integer)
viewed = Column(Integer)
created = Column(DateTime)
def __init__(self, name):
self.name = name
然后在一个视图中我试图过滤一些记录:
walls = DBSession.query(Photo).filter(Photo.cat_id == 20).limit(10)
但是这一小段代码不起作用,我出错了:
[sqlalchemy.engine.base.Engine][Dummy-2] {'param_1': 1, 'cat_id_1': 20}
*** sqlalchemy.exc.ProgrammingError: (ProgrammingError) relation "photo" does not exist
LINE 2: FROM photo
^
'SELECT photo.id AS photo_id, photo.name AS photo_name, photo.filename AS photo_filename, photo.cat_id AS photo_cat_id, photo.viewed AS photo_viewed, photo.created AS photo_created, photo.amazon_folder AS photo_amazon_folder \nFROM photo \nWHERE photo.cat_id = %(cat_id_1)s \n LIMIT %(param_1)s' {'param_1': 1, 'cat_id_1': 20}
数据库连接url正确:
sqlalchemy.url = postgres://me:pwd@localhost:5432/walls
有什么建议吗?
【问题讨论】:
-
您确定创建的 SQL 表的名称与您提供给 SQLAlchemy 的名称相同吗? PostgreSQL 中的表实际上并不称为“照片”或“照片”? (请参阅this question 上的答案)
-
我遇到了同样的错误,原因是架构和表所有者不正确,即我的情况是权限问题。
标签: sqlalchemy pyramid