使用变量作为模型查找

Question

我正在向我的视图传递一个变量，该变量是要查询的模型的名称。

model_name = 'application'
assets = model_name.objects.all()

我收到 unicode 对象没有 objects 属性的错误，这是有道理的，因为我的调试器按预期（而不是按预期）显示model_name = u'application'。

我认为这与 *args 和 **kwargs 有关（我是新手，但认为我得到了），特别是因为我的代码中的其他地方我有：

role_set = ['primary_tech', 'primary_biz', 'backup_tech', 'backup_biz']
   for role in role_set:
       records_to_change = Item.objects.filter(**{role:old_owner})

这工作正常。 我尝试了我能想到的 * 和 ** 的所有组合，以及为了保持一致性而将其包装在for model_name in [model_name]中，一切都给了我语法错误。 我错过了什么？

Python 2.7， Django 1.5

追踪：

Environment:

Request Method: GET
Request URL: http://localhost:8000/application/all/
Django Version: 1.6.1
Python Version: 2.7.2
Installed Applications:
('suit',
 'south',
 'django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.sites',
 'django.contrib.redirects',
 'django.contrib.messages',
 'django.contrib.staticfiles',
 'django.contrib.admin',
 'rest_framework',
 'ldap_sync',
 'crispy_forms',
 'ownership.apps.Catalog',
 'ownership.apps.Assets',
 'ownership.apps.Shared',
 'ownership.libs.display',
 'django_tables2',
 'haystack',
 'autocomplete_light',
 'reversion',
 'debug_toolbar')
Installed Middleware:
('django.middleware.common.CommonMiddleware',
 'django.contrib.sessions.middleware.SessionMiddleware',
 'django.middleware.csrf.CsrfViewMiddleware',
 'django.contrib.auth.middleware.AuthenticationMiddleware',
 'django.contrib.messages.middleware.MessageMiddleware',
 'django.contrib.redirects.middleware.RedirectFallbackMiddleware',
 'django.middleware.transaction.TransactionMiddleware',
 'reversion.middleware.RevisionMiddleware',
 'ownership.libs.shibboleth.CustomHeaderMiddleware',
 'ownership.libs.middleware.LoginRequiredMiddleware',
 'debug_toolbar.middleware.DebugToolbarMiddleware')

Traceback:
File "/Users/nicholsp/.virtualenvs/ownership/lib/python2.7/site-packages/django/core/handlers/base.py" in get_response
  90.                 response = middleware_method(request)
File "/Users/nicholsp/.virtualenvs/ownership/lib/python2.7/site-packages/django/utils/importlib.py" in import_module
  40.         __import__(name)
File "/Users/nicholsp/code/ownership/ownership/urls.py" in <module>
  27.     url(r'^', include('ownership.apps.Assets.urls'), name='home'),
File "/Users/nicholsp/.virtualenvs/ownership/lib/python2.7/site-packages/django/conf/urls/__init__.py" in include
  26.         urlconf_module = import_module(urlconf_module)
File "/Users/nicholsp/.virtualenvs/ownership/lib/python2.7/site-packages/django/utils/importlib.py" in import_module
  40.         __import__(name)
File "/Users/nicholsp/code/ownership/ownership/apps/Assets/urls.py" in <module>
  3. import views
Exception Type: SyntaxError at /application/all/
Exception Value: invalid syntax (views.py, line 132)

Answer 1

from django.db.models import get_model
class MyModel(models.Model):
   ...
model_class = get_model('myapp', 'mymodel')
print model_class.__name__
'MyModel'
model_class.objects.all()
[<MyModel: 1>, <MyModel: 2>, <MyModel: 3>, ... ]