尽管存在对象，但“NoneType”对象没有属性“seventh_five”

Question

我正在创建一个用户可以创建项目的网站。每个项目有 10 个宏观问题。每个宏观问题都有子问题。我正在尝试检索宏观问题的答案。projects_id包含我数据库中的所有项目。  projects = Project.objects.all()  projects_id = []  for project in projects:    projects_id.append(project.id)所以我做了一个for循环：  for project_id in projects_id:    seventh_question_project = Seventhquestion.objects.all().filter(project=project_id).first()    answer_seventh_five = seventh_question_project.seventh_five        if answer_seventh_five == "No":           unaudited_projects.append(answer_seventh_five)        else:           audited_projects.append(answer_seventh_five)如果我使用控制台，我可以检索answer_seventh_five：但是，如果加载页面，我会得到：“NoneType”对象没有属性“seventh_five”由于存在，我无法解释answer_seventh_five（因为我通过控制台检索它来测试它）可能原因是我正在过滤而不是使用get.我试过：seventh_question_project = Seventhquestion.objects.get(project=project_id)但我得到：第七题匹配查询不存在。但是： seventh_question_project = Seventhquestion.objects.all()有效

Answer 1

不要通过ID。这不是 Django ORM 的用途。只需循环浏览项目本身：

for project in Project.objects.all():

    seventh_question_project = Seventhquestion.objects.all().filter(project=project).first()  # <-- not project_id!

    # or better:

    # seventh_question_project = project.seventhquestion_set.first()  

    if seventh_question_project:

        # you should always check, because you need to write fool-proof code

        answer_seventh_five = seventh_question_project.seventh_five

        ...

但是通过关系，您可以更轻松地获取相关对象。因此，假设模型上的project字段是 a ，则相反的关系是：OneToOneFieldSeventhquestionseventhquestion

for project in Project.objects.all():

    if hasattr(project, "seventhquestion"):

        # still need to check, you never know

        answer_seventh_five = project.seventhquestion.seventh_five

        ...

Answer 2

这个怎么样？

seventh_question_project = Seventhquestion.objects.filter(project=project_id).first()

answer_seventh_five = seventh_question_project.seventh_five