具有 return 列表或字段的通用解析器的 Graphene-django
Graphene-django with generic resolver to return List or Field
我正在尝试简化我的 graphene-django 视图,以根据是否发送了参数来进行 returns graphene.List 或 graphene.Field 的单个石墨烯查询。
我正在尝试以下代码,但我不确定如何处理列表和字段响应之间的变化;
"""
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()
这是我目前的 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)
具有打开参数的输出格式不符合 graphQL 的精神。但是,您可以通过使用 employeeId
作为过滤器并将输出作为列表返回来保持相同的输出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
我正在尝试简化我的 graphene-django 视图,以根据是否发送了参数来进行 returns graphene.List 或 graphene.Field 的单个石墨烯查询。
我正在尝试以下代码,但我不确定如何处理列表和字段响应之间的变化;
"""
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()
这是我目前的 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)
具有打开参数的输出格式不符合 graphQL 的精神。但是,您可以通过使用 employeeId
作为过滤器并将输出作为列表返回来保持相同的输出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