预期字符缓冲区对象

Question

这个帖子里提到了模型。在admin.py文件中：

class GroupsAdmin(admin.ModelAdmin):

    fieldsets = [
        (None,               {'fields': ['first_year', 'many other']}),

    ]

在这个管理界面里，我可以给特别系列的汽车添加first_year。我点击“+”号，然后会弹出一个新窗口 - 在这里我可以输入年份。我输入'2011'，点击“保存”，结果却出现了错误：

TypeError: expected a character buffer object.

这很奇怪，因为上周这个功能是正常的。怎么解决这个问题呢？如果我直接去first_yearAdmin里添加年份，就没有错误。

谢谢。

编辑。错误追踪信息：

Environment:


Request Method: POST
Request URL: http://127.0.0.1:8000/admin/cars/first_year/add/?_popup=1

Django Version: 1.5
Python Version: 2.7.3
Installed Applications:
('django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.sites',
 'django.contrib.messages',
 'django.contrib.staticfiles',
 'django.contrib.sitemaps',
 'cars',
 'django_evolution',
 'django.contrib.admin')
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')


Traceback:
File "/usr/local/lib/python2.7/dist-packages/django/core/handlers/base.py" in get_response
  115.                         response = callback(request, *callback_args, **callback_kwargs)
File "/usr/local/lib/python2.7/dist-packages/django/contrib/admin/options.py" in wrapper
  372.                 return self.admin_site.admin_view(view)(*args, **kwargs)
File "/usr/local/lib/python2.7/dist-packages/django/utils/decorators.py" in _wrapped_view
  91.                     response = view_func(request, *args, **kwargs)
File "/usr/local/lib/python2.7/dist-packages/django/views/decorators/cache.py" in _wrapped_view_func
  89.         response = view_func(request, *args, **kwargs)
File "/usr/local/lib/python2.7/dist-packages/django/contrib/admin/sites.py" in inner
  202.             return view(request, *args, **kwargs)
File "/usr/local/lib/python2.7/dist-packages/django/utils/decorators.py" in _wrapper
  25.             return bound_func(*args, **kwargs)
File "/usr/local/lib/python2.7/dist-packages/django/utils/decorators.py" in _wrapped_view
  91.                     response = view_func(request, *args, **kwargs)
File "/usr/local/lib/python2.7/dist-packages/django/utils/decorators.py" in bound_func
  21.                 return func(self, *args2, **kwargs2)
File "/usr/local/lib/python2.7/dist-packages/django/db/transaction.py" in inner
  223.                 return func(*args, **kwargs)
File "/usr/local/lib/python2.7/dist-packages/django/contrib/admin/options.py" in add_view
  1010.                 return self.response_add(request, new_object)
File "/usr/local/lib/python2.7/dist-packages/django/contrib/admin/options.py" in response_add
  833.                 (escape(pk_value), escapejs(obj)))
File "/usr/local/lib/python2.7/dist-packages/django/utils/functional.py" in wrapper
  194.             return func(*args, **kwargs)
File "/usr/local/lib/python2.7/dist-packages/django/utils/html.py" in escapejs
  65.     return mark_safe(force_text(value).translate(_js_escapes))

Exception Type: TypeError at /admin/cars/first_year/add/
Exception Value: expected a character buffer object

模型

class Mark(models.Model):
    many_many_char_fields
    def __unicode__(self):
        return self.name

class First_Year(models.Model):
    year = models.IntegerField()
    def __unicode__(self):
        return str(self.year)

class Groups(models.Model):
        many_other_fields
    mark = models.ForeignKey(Mark, related_name='groups')
    last_update = models.DateTimeField()
    first_year = models.ForeignKey(First_Year, related_name='first_year')
    def __unicode__(self):
        return self.name
    def altered_date(self, year):
        altered_year = int(year)-int(self.first_year.year)
        return altered_year

Answer 1

你忘了这个

class GroupsAdmin(admin.ModelAdmin):
    model = Groups

Answer 2

其实，如果你想把year字段保持为整数类型，就没必要把它转换成字符类型。因为在很多情况下，你确实需要用到数字。针对这个1.5版本的bug，可以通过让__unicode__函数返回一个unicode字符串来解决这个问题：

class First_Year(models.Model):
    year = models.IntegerField()

    def __unicode__(self):
        return unicode(self.year)

Answer 3

在编程中，有时候我们需要处理一些数据，比如从一个地方获取数据，然后在程序中使用它。这个过程就像是从冰箱里拿食材，然后用这些食材做饭一样。

有些时候，我们会遇到一些问题，比如数据格式不对，或者数据丢失了。这就像是你打开冰箱，发现食材不够，或者有些食材坏掉了。

为了避免这些问题，我们可以提前检查一下数据，确保它们是完整和正确的。就像在做饭之前，先把所有的食材都检查一遍，确保没有坏掉的。

总之，处理数据就像做饭一样，需要仔细检查和准备，才能做出美味的菜肴。

class First_Year(models.Model):
    year = models.CharField(max_length=4)
    def __unicode__(self):
        return self.year