python从另一个调用一个函数

Question

我使用python django DRF创建了一个序列化器。我创建了一个单独的帮助器类，并尝试覆盖create方法。

我的助手班：

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

在我的序列化器中，我必须调用create方法，例如：

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，即从同一类内部的函数调用其他函数。 我对如何在函数调用中传递self有疑问。现在我无法保存模型，并且一切正常，但是在发布之后，显示create（）的实例并没有返回对象AssertionError ..但是它正在保存模型并完美地工作

Answer 1

self是对对象本身的引用。 使用self作为函数的第一个参数，您可以创建实例方法。 因此，如果要使用实例方法，则需要启动该类的对象，然后使用该对象调用该函数。 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，您可以在此处检查。

Answer 2

这工作：

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

在序列化器中：

  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