python从一个函数调用另一个函数

我正在使用python django DRF创建一个序列化程序。我正在创建一个单独的帮助器类，并尝试重写create方法

我的助手类：

class WorkFlowHelper:

    def assign_level_permissions(self,workflow_level,level):
        for permission in level['permissions']:
            workflow_permission_obj = WorkflowPermission.objects.get(short_name=permission['short_name'])
            workflow_level.permissions.add(workflow_permission_obj)

    def create_levels(self,workflow,levels):
        for level in levels:
            workflow_level = WorkflowLevel()
            workflow_level.workflow=workflow,
            workflow_level.level = level['level']
            workflow_level.operation=level['operation'] 
            workflow_level.save()
            workflow.levels.add(workflow_level)
            self.assign_level_permissions(workflow_level,level)

    def create_categories(self,workflow, categories):
       for category in categories:
           workflow_category_obj = WorkflowCategory.objects.get(short_name=category['short_name'])
           workflow.categories.add(workflow_category_obj)

    def create(self,name, description,tenant, levels, categories):
        workflow = Workflow.objects.create(name=name, description=description, tenant=tenant)
        self.create_levels(workflow,levels)
        self.create_categories(workflow,categories)
        workflow.save()

        return workflow

在序列化程序中，我必须调用create方法，如下所示：

def create(self, validated_data):
    name=validated_data['name']
    description=validated_data.get('description'),
    tenant=self.context['request'].user.tenant
    levels = self.initial_data['levels']
    categories = self.initial_data['categories']
    helper = WorkFlowHelper()
    helper.create(name,description,tenant,levels,categories)

在helper类中，我从同一类中的函数调用其他函数。我怀疑如何在函数调用中传递self。现在我可以保存模型，一切正常，但就在post之后，它的显示create没有返回对象instance AssertionError。但是它正在保存模型，并且工作得很好。

self是对对象本身的引用。使用self作为函数的第一个参数，可以创建实例方法。因此，如果要使用实例方法，则需要初始化该类的对象，然后使用该对象调用函数。例如，在这里：

helper = WorkFlowHelper()  # create a `WorkFlowHelper` object

helper.create(creates,name,description,tenant,levels,categories)
# here you don't need to pass `self`, because it is referencing to it to the helper object

作为参考，您可以检查。

此操作有效：

class WorkFlowHelper:


    def assign_level_permissions(self,workflow_level,level):
        for permission in level['permissions']:
            workflow_permission_obj = WorkflowPermission.objects.get(short_name=permission['short_name'])
            workflow_level.permissions.add(workflow_permission_obj)

    def create_levels(self,workflow,levels):
        for level in levels:
            workflow_level = WorkflowLevel()
            workflow_level.workflow=workflow,
            workflow_level.level = level['level']
            workflow_level.operation=level['operation'] 
            workflow_level.save()
            workflow.levels.add(workflow_level)
            self.assign_level_permissions(workflow_level,level)

    def create_categories(self,workflow, categories):
       for category in categories:
           workflow_category_obj = WorkflowCategory.objects.get(short_name=category['short_name'])
           workflow.categories.add(workflow_category_obj)

    def create(self,name, description,tenant, levels, categories):
        workflow = Workflow.objects.create(name=name, description=description, tenant=tenant)
        self.create_levels(workflow,levels)
        self.create_categories(workflow,categories)
        workflow.save()

        return workflow

在序列化程序中：

  def create(self, validated_data):
        name=validated_data['name']
        description=validated_data.get('description'),
        tenant=self.context['request'].user.tenant
        levels = self.initial_data['levels']
        categories = self.initial_data['categories']
        helper = WorkFlowHelper()
        workflow = helper.create(name,description,tenant,levels,categories)

        return workflow

---

你太难了。当您使用“self”从一个方法函数引用另一个方法函数时，该函数已将“self”作为第一个参数传递。因此，您将“工作流”作为第一个参数传递的所有位置都是错误的。我不知道仅仅把它们拿出来是否能给你你想要的逻辑，但是电话会很好地组织起来。比如：self.create\u categories工作流，categories->self.create\u categories categories你能检查我编辑的问题吗@乔德明