django - django db ProgrammingError
问题描述
当我尝试重置数据库时(我创建了一个新的空数据库)
当我运行这个命令python manage.py
makemigrations(或迁移)时出现这个错误
for pattern in self.url_patterns:
File "django\utils\functional.py", line 48, in __get__
res = instance.__dict__[self.name] = self.func(instance)
File "django\urls\resolvers.py", line 589, in url_patterns
patterns = getattr(self.urlconf_module, "urlpatterns", self.urlconf_module)
File "django\utils\functional.py", line 48, in __get__
res = instance.__dict__[self.name] = self.func(instance)
File "django\urls\resolvers.py", line 582, in urlconf_module
return import_module(self.urlconf_name)
File \python\python39\lib\importlib\__init__.py", line 127, in import_module
return _bootstrap._gcd_import(name[level:], package, level)
File "<frozen importlib._bootstrap>", line 1030, in _gcd_import
File "<frozen importlib._bootstrap>", line 1007, in _find_and_load
File "<frozen importlib._bootstrap>", line 986, in _find_and_load_unlocked
File "<frozen importlib._bootstrap>", line 680, in _load_unlocked
File "<frozen importlib._bootstrap_external>", line 790, in exec_module
File "<frozen importlib._bootstrap>", line 228, in _call_with_frames_removed
File "project\urls.py", line 18, in <module>
path('api/v2/', include('project.api.urls'), name='api'),
File "django\urls\conf.py", line 34, in include
urlconf_module = import_module(urlconf_module)
File "python\python39\lib\importlib\__init__.py", line 127, in import_module
return _bootstrap._gcd_import(name[level:], package, level)
File "<frozen importlib._bootstrap>", line 1030, in _gcd_import
File "<frozen importlib._bootstrap>", line 1007, in _find_and_load
File "<frozen importlib._bootstrap>", line 986, in _find_and_load_unlocked
File "<frozen importlib._bootstrap>", line 680, in _load_unlocked
File "<frozen importlib._bootstrap_external>", line 790, in exec_module
File "<frozen importlib._bootstrap>", line 228, in _call_with_frames_removed
File "project\api\urls.py", line 3, in <module>
from .views import *
File "project\api\views.py", line 146, in <module>
class SampleViewSet(viewsets.ModelViewSet):
File "project\api\views.py", line 148, in SampleViewSet
general = General.objects.get(is_active=True)
File "django\db\models\manager.py", line 85, in manager_method
return getattr(self.get_queryset(), name)(*args, **kwargs)
File "django\db\models\query.py", line 425, in get
num = len(clone)
File "django\db\models\query.py", line 269, in __len__
self._fetch_all()
File "django\db\models\query.py", line 1308, in _fetch_all
self._result_cache = list(self._iterable_class(self))
File "django\db\models\query.py", line 53, in __iter__
results = compiler.execute_sql(chunked_fetch=self.chunked_fetch, chunk_size=self.chunk_size)
File "django\db\models\sql\compiler.py", line 1156, in execute_sql
cursor.execute(sql, params)
File "django\db\backends\utils.py", line 98, in execute
return super().execute(sql, params)
File "django\db\backends\utils.py", line 66, in execute
return self._execute_with_wrappers(sql, params, many=False, executor=self._execute)
File "django\db\backends\utils.py", line 75, in _execute_with_wrappers
return executor(sql, params, many, context)
File "django\db\backends\utils.py", line 84, in _execute
return self.cursor.execute(sql, params)
File "env\lib\site-packages\django\db\utils.py", line 90, in __exit__
raise dj_exc_value.with_traceback(traceback) from exc_value
File "env\lib\site-packages\django\db\backends\utils.py", line 84, in _execute
return self.cursor.execute(sql, params)
django.db.utils.ProgrammingError: HATA: "app_general" object doesnt exist
LINE 1: ...", "app_general"."field1" FROM "field_2...
视图.py
class SampleViewSet(viewsets.ModelViewSet):
general = General.objects.get(is_active=True)
queryset = model.objects.filter(field=site_general.field)
我删除了所有应用程序中的所有迁移文件(没有 __init__
)以及 .pyc 文件
这个问题有什么想法吗?
(编辑我添加了 Traceback)
解决方案
类的所有非方法属性在类的声明或定义期间被评估/执行。因此,当解释器到达您的类时SampleViewSet
,它会尝试执行该行general = General.objects.get(is_active=True)
,但您的数据库当然还不存在,并且由于调用.get
意味着数据库查询,您会收到错误。
一般来说,任何可能导致数据库查询的语句都应该由应用程序配置ready
方法、某些中间件/上下文处理器或请求中的视图运行。查看代码,它看起来不像你general
在任何地方使用,所以你可以简单地删除它:
class SampleViewSet(viewsets.ModelViewSet):
queryset = model.objects.filter(field=site_general.field)
您可能会说model.objects.filter(...)
也会导致查询,但.filter
返回 aQuerySet
和查询集是惰性的,因此仅在需要时才进行评估,因此不会导致任何错误。
推荐阅读
- javascript - 向下滚动时如何隐藏导航菜单的图像?
- python - 如何序列化从张量流数组创建的字典对象
- twitter-bootstrap - 单击后,工具提示对 btn-group-toggle 中的单选按钮保持可见
- oracle - OBIEE 计数不考虑应用的过滤器
- android - 推送失败:无法访问 URL:请求的 URL 返回错误:409
- java - 弹簧反应器中的错误处理未按预期工作
- corda - 我什么时候需要 waitForLedgerCommit
- azure - 主动/被动模式下的 Azure 2 负载均衡器
- amazon-web-services - 如果 VPC 中的 lambda 函数可以在没有 AWSLambdaVPCAccessExecutionRole 的情况下访问同一 VPC 中的弹性搜索
- java - 不能在片段内显示对话框?