小编D_D*_*D_D的帖子

我想为我的django项目生成一个基本的数据库模式,用模型和模型字段显示我的所有应用程序和边界条件等.在python中是否已经有任何django的数据库模式生成器？或者我应该怎么做呢.

我试图在我的视图中覆盖get()方法:

broadcast = Broadcast.objects.get(request, pk = broadcast_id) 

在我的模型中,重写方法如下:

class Broadcast(models.Model):      
    person = models.ForeignKey(User)
    post = models.CharField(max_length=300 , verbose_name = 'Say it out loud !')
    .
    .

    def get(self, *args, **kwargs):
        if request.user == self.person :
            super(Broadcast, self).get(*args, **kwargs)
        else :
            return none

两个问题 - >
我最重要的地方出错了？
如何访问传入get的"请求"参数？

我创建了以下模型

 class ConnectToFrom(models.Model):
    person = models.ForeignKey(User, null=True)
    kiosk = models.ForeignKey(Kiosks, null=True)

 class Connect(models.Model):
    parent_id = models.ForeignKey("self", null = True, blank = True)
    sender = models.ForeignKey(ConnectToFrom, related_name='sent_messages' )
    reciever = models.ForeignKey(ConnectToFrom, related_name='received_messages')
    .
    .
    .

我无法通过管理站点访问连接以添加任何连接对象!我不能将相同的模型引用到两个字段吗？我无法弄清楚导致错误的确切原因.请帮忙

追溯 :

Environment:


Request Method: GET
Request URL: http://www.krisvenham.com:8000/admin/connect/connect/

Django Version: 1.3
Python Version: 2.6.6
Installed Applications:
['broadcast',
 'shastra',
 'fb_api',
 'log',
 'nties',
 'crc',
 'connect',
 'network',
 'django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.sites',
 'django.contrib.messages',
 'django.contrib.admin',
 'kiosks',
 'content',
 'home',
 'dashboard',
 'trial',
 'meta',
 'oembed']
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') …