对嵌套字段graphql中的结果列表进行排序 - python - grapheneSQLAlchemy答案

【问题标题】：Sort list of results in nested field graphql - python - grapheneSQLAlchemy对嵌套字段graphql中的结果列表进行排序 - python - grapheneSQLAlchemy
【发布时间】：2022-06-27 04:02:47
【问题描述】：

我有一个关于如何使用 GraphQL 以多对一关系对结果列表进行排序的方式的问题。让我们以graphene-sqlalchemy的例子为例。

架构是：

class Department(Base):
    __tablename__ = 'department'
    id = Column(Integer, primary_key=True)
    name = Column(String)


class Employee(Base):
    __tablename__ = 'employee'
    id = Column(Integer, primary_key=True)
    name = Column(String)
    hired_on = Column(DateTime, default=func.now())
    department_id = Column(Integer, ForeignKey('department.id'))
    department = relationship(
        Department,
        backref=backref('employees',
                    uselist=True,
                    cascade='delete,all'))

这是我的架构：

    import graphene
    from models import (Employee as EmployeeModel, Department as DepartmentModel)
    from graphene_sqlalchemy import (
        SQLAlchemyObjectType
    )
    
    class Employee(SQLAlchemyObjectType):
        class Meta:
            model = EmployeeModel


    class Department(SQLAlchemyObjectType):
        class Meta:
            model = DepartmentModek

    class Query(graphene.ObjectType):
        find_departments = graphene.List(of_type = Department)
        def resolve_find_departments(root, info) : 
            return db.session.query(DepartmentModel).all()

在我的查询中，我想访问我部门的员工列表并按名称对结果进行排序。我的查询可能如下所示：

     query findDepartments(){
        findDepartments{
            department{
                employee(orderBy : 'name'){
                    name
                }
            }
    
        }
    }

但是这个查询不起作用。如何实现嵌套字段员工的排序？

【问题讨论】：

标签： sqlalchemy graphql graphene-python

【解决方案1】：

有多种方法可以做到这一点，但一种简单的方法是创建一个具有您想要的排序的自定义解析器，即类似

class Query(graphene.ObjectType):
    find_departments = graphene.List(of_type = Department)
    find_departments_by_name = graphene.List(of_type = Department)
    
   def resolve_find_departments(root, info): 
        return db.session.query(DepartmentModel).all()

    def resolve_find_departments_by_name(root, info): 
        return db.session.query(DepartmentModel).order_by("name").all()  # this is a guess of how sqlalchemy sorting syntax

【讨论】：

这里的问题是我想按名称而不是部门来排序员工。 Employee 是 DepartmentModel 中的一个嵌套字段。
明白了。此答案不适用于您的需要。 @jorzel 的答案似乎在正确的轨道上（尽管它需要缩进修复。）

【解决方案2】：

您可以覆盖 employees 的解析器：

    import graphene
    from models import (Employee as EmployeeModel, Department as DepartmentModel)
    from graphene_sqlalchemy import (
        SQLAlchemyObjectType
    )
    
    class Employee(SQLAlchemyObjectType):
        class Meta:
            model = EmployeeModel


    class Department(SQLAlchemyObjectType):
        class Meta:
            model = DepartmentModel
    
        employees = graphene.List(of_type=Employee, order_by=graphene.String())
        def resolve_employees(root, info, **kwargs):
            query = self.employees
            order_arg = kwargs.get('order_by')
            if order_org:
                # should check that `order_arg` is a column of `Employee`
                query = query.order_by(order_arg)
            return query.all()
       

    class Query(graphene.ObjectType):
        find_departments = graphene.List(of_type = Department)
        def resolve_find_departments(root, info) : 
            return db.session.query(DepartmentModel).all()

【讨论】：