python - SQLAlchemy 一个(): No Results Found

Question

我不断收到 Traceback 提示系统找不到结果。鉴于对象实际上正在经历，情况不应该如此(我知道这一点，因为我有一个打印命令来确认查询正在经历什么)。我相信问题出在 create_category_rating(..) 方法中。

当我使用 .one() 时，它会引发下面的投诉。当我使用 .all 时，它返回 AttributeError: 'list' object has no attribute '_sa_instance_state' (其他人有这个 issue )。 docs SQLAlchemy 很好地定义了这些方法，但由于某种原因我的代码崩溃了。

问题:如果对象正在通过检索方法，是什么导致程序中断？某处有一个错误，我一直在努力解决它。

我正在使用 SQLAlchemy 和 python2.7:

回溯:

ERROR: notssdb.test.test.test1
----------------------------------------------------------------------
Traceback (most recent call last):
  File "/usr/local/lib/python2.7/site-packages/nose/case.py", line 197, in runTest
    self.test(*self.arg)
  File "/Users/ack/code/venv/NotssDB/notssdb/test/test.py", line 111, in test1
    api.create_category_rating(2, 'Decision-Making', 'baseball', 'Becoming a Leader')
  File "/Users/ack/code/venv/NotssDB/notssdb/api/convenience.py", line 41, in create_category_rating
    assessment_results = self.retrieve_assessment_results(owner, assessment)
  File "/Users/ack/code/venv/NotssDB/notssdb/api/object.py", line 324, in retrieve_assessment_results
    filter(Assessment_Results.owner == owner).one()
  File "/usr/local/lib/python2.7/site-packages/sqlalchemy/orm/query.py", line 2478, in one
    raise orm_exc.NoResultFound("No row was found for one()")
NoResultFound: No row was found for one()
   # output from Traceback 
   Decision-Making
   baseball
   Becoming a Leader
   Becoming a Leader
   test1 <Assessment(name='Becoming a Leader', text='better decisions')>

便利.py(创建方法):

def create_category_rating(self, category_rating_int, category_name, username, name):
    category = self.retrieve_category(category_name)
    owner = self.retrieve_user(username) # added
    assessment = self.retrieve_assessment(name) #added
    assessment_results = self.retrieve_assessment_results(owner, assessment)
    return super(ConvenienceAPI, self).create_category_rating(category_rating_int, category, assessment_results)

对象.py(检索方法):

def retrieve_assessment_results(self, *args):
    id, assessment, owner = None, None, None
    if len(args) == 1:
        id, = args[0]
    elif len(args) == 2:
        assessment, owner = args
    else:
        raise ValueError('Value being passed is an object')
    if id is not None:
        return self.session.query(Assessment_Results).\
        filter(Assessment_Results.id == id).one()
    elif owner is not None:
        print 'test1', owner
        return self.session.query(Assessment_Results).\
        filter(Assessment_Results.owner == owner).one()
    elif assessment is not None:
        print 'test2', assessment
        return self.session.query(Assessment_Results).\
        filter(Assessment_Results.assessment == assessment).one()

Answer 1

您有以下内容:

self.retrieve_assessment_results(owner, assessment)

...但是“retrieve_assessment_results”的参数如下(添加了我的评论)...

def retrieve_assessment_results(self, *args):
    id, assessment, owner = None, None, None
    if len(args) == 1:
        id, = args[0]
    elif len(args) == 2:
        assessment, owner = args # Note this line!
    else:
        raise ValueError('Value being passed is an object')

你通过了所有者，然后是评估。但是如果你检测到两个对象，你会读评估，然后是所有者。换句话说，您正在使用一种对象顺序调用该方法，并假设相反的顺序进行检索。您甚至可以在调试行中看到这种情况:

print 'test1', owner

你会期望这会打印类似 test1 <User(...)> 的内容, 但它打印了...

test1 <Assessment(name='Becoming a Leader', text='better decisions')>

另外，考虑使用 Python keyword arguments相反:

def retrieve_assessment_results(self, id=None, assessment=None, owner=None):
    ...

然后你的电话就变得不那么模棱两可了:

self.retrieve_assessment_results(owner=owner, assessment=assessment)