带有通用解析器的 Graphene-django 返回列表或字段

Question

我正在尝试简化我的 graphene-django 视图，使其具有单个石墨烯查询，该查询根据是否发送参数返回一个 graphene.List 或一个 graphene.Field。

我正在尝试以下代码，但我不确定如何处理 List 和 Field 响应之间的变化；

    """ 
        graphene.Field & graphene.List will not be determined until the resolver 'resolve_employee' 
        checks if a employeeId param is sent.

        Issue : How can I make this generic to return a list or a field
    """


    employee = graphene.Field(EmployeeType, employeeId=graphene.Int())

    def resolve_employee(self, info, **kwargs):
        employeeId = kwargs.get('employeeId')
        if employeeId is not None:
            return Employee.objects.get(pk=employeeId)
        else:
            return Employee.objects.all()

这是我当前的 schema.py，有两个单独的

class EmployeeType(DjangoObjectType):
    class Meta:
        model = Employee


class Query(object):)
    allEmployees = graphene.List(EmployeeType, active=graphene.Boolean())
    employeeDetail = graphene.Field(EmployeeType, employeeId=graphene.Int())

        
    def resolve_allEmployees(self, info, **kwargs):
        active_param = kwargs.get('active')
        if type(active_param) == bool:
            return Employee.objects.filter(term_date__isnull=active_param)
        return Employee.objects.all()


    def resolve_employeeDetail(self, info, **kwargs):
        employeeId = kwargs.get('employeeId')
        if employeeId is not None:
            return Employee.objects.get(pk=employeeId)

Answer 1

具有打开参数的 output 格式不符合 graphQL 的精神。 但是，您可以通过使用employeeId作为过滤器并返回 output 作为列表来保持相同的 output 格式来解决您的问题。 例如：

def resolve_Employees(self, info, **kwargs):
    active_param = kwargs.get('active')
    employee_id = kwargs.get('employeeId')
    employees = Employees.objects.all()
    if type(active_param) == bool:
        employees = employees.filter(term_date__isnull=active_param)
    if employee_id:
        employees =  employees.filter(id=employee_id)  # filter list to just one employee
    return employees