Flask Untivent-关系的搜索筛选器。属性！=一对多关系中的val

我的书和行动表的炼金术模型如下

class Book(db.Model):

    __tablename__ = 'book'

    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(255))
    activity = db.relationship('Actions')

class Actions(db.Model):

    __tablename__ = 'actions'

    id = db.Column(db.Integer, primary_key=True)
    book_id = db.Column(db.Integer, db.ForeignKey('book.id'), nullable=False)
    action = db.Column(db.Enum(['requested', 'issued', 'opened', 'closed']), nullable=True)
    created_on = db.Column(db.DateTime, default=_get_date_time)

基本上，这是一本书和行动之间的一对多关系。我使用的是API

我需要去

所有尚未关闭的书籍

我在下面试过

及

但我得到了错误的结果

{
"num_results": 30,
"objects": [
{
  "activity": [
    {
      "action": "opened",
      "created_on": "2015-06-05T17:05:07",
      "id": 31
    },
    {
      "action": "closed",
      "created_on": "2015-06-05T17:05:44",
      "id": 32
    }
  ],
  "id": 1,
  "name": "K&R"
  ....
},
...
]
"page": 1,
"total_pages": 3
}

我在这里犯了什么错误，还是这种事情在不安中是不可能发生的

请帮帮我。

试试这个：

q = { "filters":[   
         {
          "name":"activity__action", 
          "op":"not_equal_to", 
          "val":"closed"
         }
      ]
}

---

它给出了一个错误-无法将集合与对象或集合进行比较；使用contains测试成员资格。当response={message:cannot constructure query}且op:any时，val:opened？但最后一段执行类似于从存在的书本中选择*，从存在的书本中选择1，其中book.id=actions.book\u id和actions.action=%s ORDER BY book.id asc，我想我需要类似于从不存在的书本中选择*，从存在的书本中选择1，其中book.id=actions.book\u id和actions.action='closed'按book.id订购ASC@Hussain你找到解决办法了吗？

{
"num_results": 30,
"objects": [
{
  "activity": [
    {
      "action": "opened",
      "created_on": "2015-06-05T17:05:07",
      "id": 31
    },
    {
      "action": "closed",
      "created_on": "2015-06-05T17:05:44",
      "id": 32
    }
  ],
  "id": 1,
  "name": "K&R"
  ....
},
...
]
"page": 1,
"total_pages": 3
}

q = { "filters":[   
         {
          "name":"activity__action", 
          "op":"not_equal_to", 
          "val":"closed"
         }
      ]
}