Sqlalchemy 炼金术。如何将字段打包回查询中?

Sqlalchemy 炼金术。如何将字段打包回查询中?,sqlalchemy,graphql,graphene-python,Sqlalchemy,Graphql,Graphene Python,为了满足我的需要,我使用了表间连接。在循环中,我需要检查条件的相关性,并在执行时将查询添加到最终选择中。请告诉我如何将满足条件的当前查询添加到“答案”选择中 这是我的密码: def resolve_doops(self,info): answer=[] query = db.query(DOOP,Direction,Category,Category_DOOP) query = query.join(Direction, Direction.

为了满足我的需要,我使用了表间连接。在循环中,我需要检查条件的相关性,并在执行时将查询添加到最终选择中。请告诉我如何将满足条件的当前查询添加到“答案”选择中

这是我的密码:

   def resolve_doops(self,info):
        answer=[]
        query = db.query(DOOP,Direction,Category,Category_DOOP)
        query = query.join(Direction, Direction.id_cluster == DOOP.id_cluster)
        query = query.join(Category_DOOP, Category_DOOP.id_doop == DOOP.id)
        query = query.join(Category, Category_DOOP.id_category == Category.id)
        query = query.all()

        for doop,diretion,category_doop,category in query:
            if (doop.ovz == bool(self.ovz)):
                answer.append(????)
        return answer
Upd(我发现了一个更优雅的解决方案):


现在还不太清楚你想在这里做什么-你想做
回答.附加((doop,diretion,category\u doop,category))
?如果不是,你认为“当前查询”是什么意思?@snakecharmerb谢谢你的回答!它对我有效:answer.append((doop,diretion,category_doop,category)),但我可以从另一个角度来处理它,在你发表评论之前,我已经修改了这个方法并创建了一个已经启用过滤的查询。
 def resolve_doops(self,info):

        query = db.query(DOOP,Direction,Category,Category_DOOP)
        query = query.filter_by(ovz=self.ovz)
        query = query.join(Direction, Direction.id_cluster == DOOP.id_cluster)
        query = query.filter(Direction.name.in_(self.direction))
        query = query.join(Category_DOOP, Category_DOOP.id_doop == DOOP.id)
        query = query.join(Category, Category_DOOP.id_category == Category.id)
        query = query.filter(Category.age_max.in_(self.age),Category.age_min.in_(self.age))
        query = query.all()

        return query