我正在創建一個用戶可以創建項目的網站。每個項目有 10 個宏觀問題。每個宏觀問題都有子問題。我正在嘗試檢索宏觀問題的答案。projects_id包含我數據庫中的所有項目。 projects = Project.objects.all() projects_id = [] for project in projects: projects_id.append(project.id)所以我做了一個for循環： 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)如果我使用控制臺，我可以檢索answer_seventh_five：但是，如果加載頁面，我會得到：“NoneType”對象沒有屬性“seventh_five”由于存在，我無法解釋answer_seventh_five（因為我通過控制臺檢索它來測試它）可能原因是我正在過濾而不是使用get.我試過：seventh_question_project = Seventhquestion.objects.get(project=project_id)但我得到：第七題匹配查詢不存在。但是： seventh_question_project = Seventhquestion.objects.all()有效
2 回答
慕虎7371278
TA貢獻1802條經驗 獲得超4個贊
不要通過ID。這不是 Django ORM 的用途。只需循環瀏覽項目本身：
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
...
但是通過關系，您可以更輕松地獲取相關對象。因此，假設模型上的project字段是 a ，則相反的關系是：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
...
縹緲止盈
TA貢獻2041條經驗 獲得超4個贊
這個怎么樣？
seventh_question_project = Seventhquestion.objects.filter(project=project_id).first()
answer_seventh_five = seventh_question_project.seventh_five
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