Question

我制作了一个简单的Django应用程序。我有一个模型“访客”。我的目标是在Django管理员中出现两个两个表。一个与所有访客一起，一个只有今天的访客。

按照these instructions的说法，我完成了以下代码。但是我的麻烦是，在管理页面中编辑组时，我无法让VisitorExpectedTodayProxy显示在“可用权限”中。有谁知道怎么做？

Models.py

class Visitor(models.Model):
    visit_datetime = models.DateTimeField(null=True)
    visitor_name = models.CharField(max_length=500)

#Make dummy models for different object views in admin interface
class VisitorExpectedTodayProxy(Visitor):
    class Meta:
        proxy=True
        verbose_name = "Visitor"
        verbose_name_plural = "Today's Visitors and Regular Visitors"

更新

我确实运行了syncdb，但我仍然没有在管理网站上看到它。 syncdb的结果：

$ python manage.py syncdb
Syncing...
No fixtures found.

Synced:
 > django.contrib.auth
 > django.contrib.contenttypes
 > django.contrib.sessions
 > django.contrib.sites
 > django.contrib.messages
 > django.contrib.admin
 > south

Answer 1

这是我用来手动输入代理对象的ContentType条目的脚本，然后为它们创建新的权限。

这只是生成sql，然后你应该在mysql中运行。之后，它们应显示在用户的权限列表中，并且需要添加它们。这些的一些命名约定并不明显。

your_models=['proxy model',
            ]

for model in models:
    model_nospace = model.replace(' ','')
    sql = 'insert into django_content_type (name, app_label, model) values     ("%s","<<app_name>>","%s");'%(model,model_nospace)
    print sql
    for kind, permname in [('Can add','add_%s'%model_nospace),
                          ('Can change','change_%s'%model_nospace),
                          ('Can delete','delete_%s'%model_nospace),]:
        sql = 'insert into auth_permission (name, content_type_id, codename) values ("%s     %s",(select id from django_content_type where name="%s"),"%s");'%    (kind,model,model,permname)
        print sql

Answer 2

我意识到这个问题已经暂时关闭了，但是我会分享对我有用的东西，以防它可以帮助别人。

事实证明，即使我创建的代理模型的权限列在syncdb之后的父应用下，即使我授予非超级用户所有权限，但仍然拒绝访问我的代理模特通过管理员。

虽然我还没有尝试过基于SQL的修复程序，但修复Django级别的bug对我有用。您必须解决已知的Django错误（https://code.djangoproject.com/ticket/11154）并连接到post_syncdb信号以正确创建代理模型的权限。以下代码根据该主题的一些评论从https://djangosnippets.org/snippets/2677/修改。

我把它放在myapp / models.py中，它包含我的代理模型。从理论上讲，这可以存放在INSTALLED_APPS之后的任何django.contrib.contenttypes中，因为需要在update_contenttypes处理程序注册post_syncdb信号后加载，因此我们可以将其断开连接。< / p>

def create_proxy_permissions(app, created_models, verbosity, **kwargs):
    """
    Creates permissions for proxy models which are not created automatically
    by 'django.contrib.auth.management.create_permissions'.
    See https://code.djangoproject.com/ticket/11154
    Source: https://djangosnippets.org/snippets/2677/

    Since we can't rely on 'get_for_model' we must fallback to
    'get_by_natural_key'. However, this method doesn't automatically create
    missing 'ContentType' so we must ensure all the models' 'ContentType's are
    created before running this method. We do so by un-registering the
    'update_contenttypes' 'post_syncdb' signal and calling it in here just
    before doing everything.
    """
    update_contenttypes(app, created_models, verbosity, **kwargs)
    app_models = models.get_models(app)
    # The permissions we're looking for as (content_type, (codename, name))
    searched_perms = list()
    # The codenames and ctypes that should exist.
    ctypes = set()
    for model in app_models:
        opts = model._meta
        if opts.proxy:
            # Can't use 'get_for_model' here since it doesn't return
            # the correct 'ContentType' for proxy models.
            # See https://code.djangoproject.com/ticket/17648
            app_label, model = opts.app_label, opts.object_name.lower()
            ctype = ContentType.objects.get_by_natural_key(app_label, model)
            ctypes.add(ctype)
            for perm in _get_all_permissions(opts, ctype):
                searched_perms.append((ctype, perm))

    # Find all the Permissions that have a content_type for a model we're
    # looking for. We don't need to check for codenames since we already have
    # a list of the ones we're going to create.
    all_perms = set(Permission.objects.filter(
        content_type__in=ctypes,
    ).values_list(
        "content_type", "codename"
    ))

    objs = [
        Permission(codename=codename, name=name, content_type=ctype)
        for ctype, (codename, name) in searched_perms
        if (ctype.pk, codename) not in all_perms
    ]
    Permission.objects.bulk_create(objs)
    if verbosity >= 2:
        for obj in objs:
            sys.stdout.write("Adding permission '%s'" % obj)


models.signals.post_syncdb.connect(create_proxy_permissions)
# See 'create_proxy_permissions' docstring to understand why we un-register
# this signal handler.
models.signals.post_syncdb.disconnect(update_contenttypes)

Answer 3

 ./manage.py syncdb

多数民众赞成。它将条目添加到auth_permission表中。

Answer 4

我做的是解决这个问题，我删除了每个表（仅删除auth表是不够的）。然后我从settings.py中已安装的应用程序向南注释（不确定是否会导致问题）。

然后我运行了manage.py syncdb，现在一切正常。

我只需要重新加载我的所有数据。我还不确定它是如何搞砸的。

Answer 5

I documented my hack to work around this here.

Django - 如何使ModelAdmin类显示在可用权限中

5 个答案: