Python錯誤：未定義名稱'admin'

Question

我正在第一次在Django中創建一個Python應用程序。我知道我必須取消對urls.py中的管理工具的評論，我已經這樣做了。我還添加了autodiscover。每次我嘗試了新的功能添加到管理面板，我得到這個錯誤：

「NameError：名‘admin’沒有定義」

這裏是我用我的模型添加到代碼管理面板：

class ChoiceInline(admin.StackedInline): 
    model = Choice 
    extra = 3 

    class PollAdmin(admin.ModelAdmin): 
    fieldsets = [ 
     (None,    {'fields': ['question']}), 
     ('Date information', {'fields': ['pub_date'], 'classes': ['collapse']}), 
    ] 
    inlines = [ChoiceInline] 

這裏是在Python端的代碼，我使用

admin.site.register(Poll, PollAdmin) 

，這裏是從我的urls.py代碼：

from django.conf.urls import patterns, include, url 

# Uncomment the next two lines to enable the admin: 
from django.contrib import admin 
admin.autodiscover() 

urlpatterns = patterns('', 
    # Examples: 
    # url(r'^$', 'iFriends.views.home', name='home'), 
    # url(r'^iFriends/', include('iFriends.foo.urls')), 

    # Uncomment the admin/doc line below to enable admin documentation: 
    # url(r'^admin/doc/', include('django.contrib.admindocs.urls')), 

    # Uncomment the next line to enable the admin: 
    url(r'^admin/', include(admin.site.urls)), 
    ) 

任何人都知道爲什麼它不能找到管理員名稱？

編輯

這是我整個模型文件：

from django.db import models 

class Poll(models.Model): 
    question = models.CharField(max_length=200) 
    pub_date = models.DateTimeField('date published') 
    def __unicode__(self): 
     return self.question 
    def was_published_recently(self): 
     return self.pub_date >= timezone.now() - datetime.timedelta(days=1) 

    class Choice(models.Model): 
    poll = models.ForeignKey(Poll) 
    choice_text = models.CharField(max_length=200) 
    votes = models.IntegerField() 
    def __unicode__(self): 
     return self.choice_text 


#COMMENTED OUT UNTIL I FIX THE ADMIN NAME 
from django.config import admin 

class ChoiceInline(admin.StackedInline): 
    model = Choice 
    extra = 3 

    class PollAdmin(admin.ModelAdmin): 
    fieldsets = [ 
     (None,    {'fields': ['question']}), 
     ('Date information', {'fields': ['pub_date'], 'classes': ['collapse']}), 
    ] 
    inlines = [ChoiceInline] 

#ADD THIS TO THE MAIN PYTHON FUNCTION 
admin.site.register(Poll, PollAdmin) 

Answer 1

我懷疑你沒有from django.contrib import admin在你定義ModelAdmin類，class ChoiceInline(admin.StackedInline):模型文件。

Answer 2

你應該把管理員密碼（#COMMENTED之後一切，直到我定的管理員名稱），以admin.py的

Answer 3

from django.config import admin應該from django.contrib import admin

Answer 4

你**url.py**文件的頂部，添加以下代碼

from django.contrib import admin 
admin.autodiscover() 

所以特定塊應該看起來像下面

---

from django.conf.urls import patterns, include, url 
**from django.contrib import admin 
admin.autodiscover()** 

# Uncomment the next two lines to enable the admin: 
# from django.contrib import admin 
# admin.autodiscover() 

urlpatterns = patterns('', 
    # Examples: 
    # url(r'^$', 'examplesite.views.home', name='home'), 
    # url(r'^examplesite/', include('examplesite.foo.urls')), 

    # Uncomment the admin/doc line below to enable admin documentation: 
    #url(r'^admin/doc/', include('django.contrib.admindocs.urls')), 

    # Uncomment the next line to enable the admin: 
**url(r'^admin/', include(admin.site.urls)),** 

) 

Answer 5

經過漫長而痛苦的追求解決這個愚蠢的問題，我終於遇到了答案！另一個Django程序員遇到了同樣的問題，並發現：

在ChoiceInLine的父親（本教程中您將看到「admin.StackedInline」），L的StackedInline不應該被大寫......它是就這麼簡單......三江源這麼多克倫特雷西...

http://grokbase.com/p/gg/django-users/133v4wx0sx/django-1-5-tutorial-error-module-has-no-attribute-stackedinline

Answer 6

我改變我的urls.py這樣的...這應該是基本的格式

from django.conf.urls import include, url 
from django.contrib import admin 

admin.autodiscover() 
urlpatterns = [ 
    url(r'^admin/', include(admin.site.urls)), 
    ]