python django ORM 错误：无法将关键字 'id' 解析到字段中。选择项：id，

Question

我找到了答案！这个问题发生是因为我把脚本的名字和应用的名字起得一模一样！

这是我对帖子内容的简化和修正版本。

我正在尝试运行这个独立的脚本（或者说它就在应用的文件夹里）：

#!/usr/local/bin/python -W ignore
# coding: utf-8

import sys, os
sys.path.append('/usr/home/code')
os.environ['DJANGO_SETTINGS_MODULE'] = 'tuppy.settings'
from tuppy.tup.models import *

some_dict={}
print UserProfile.objects.filter(id=1)
print 'lallala'
print some_dict['unexisting_key']

然后我遇到了以下错误。请注意，这个脚本一开始会正确地把请求结果打印到UserProfile上，但在遇到另一个错误时，它才会打印出错误的描述：

# ./tup.py
[<UserProfile: 115>]
lallala
Traceback (most recent call last):
  File "./tup.py", line 10, in <module>
    p = UserProfile.objects.filter(id=1)
  File "/usr/local/lib/python2.6/site-packages/django/db/models/manager.py", line 141, in filter
    return self.get_query_set().filter(*args, **kwargs)
  File "/usr/local/lib/python2.6/site-packages/django/db/models/query.py", line 556, in filter
    return self._filter_or_exclude(False, *args, **kwargs)
  File "/usr/local/lib/python2.6/site-packages/django/db/models/query.py", line 574, in _filter_or_exclude
    clone.query.add_q(Q(*args, **kwargs))
  File "/usr/local/lib/python2.6/site-packages/django/db/models/sql/query.py", line 1152, in add_q
    can_reuse=used_aliases)
  File "/usr/local/lib/python2.6/site-packages/django/db/models/sql/query.py", line 1045, in add_filter
    negate=negate, process_extras=process_extras)
  File "/usr/local/lib/python2.6/site-packages/django/db/models/sql/query.py", line 1215, in setup_joins
    "Choices are: %s" % (name, ", ".join(names)))
django.core.exceptions.FieldError: Cannot resolve keyword 'id' into field. Choices are: credit_limit, id, insured_order_limit, mob_tel, resale_limit, sec_tel, status, user, voice_psw

#

Answer 1

我在尝试从auth_group获取第一个组的时候也遇到了同样的问题（Django版本1.3.5）。

Group.objects.get(name='First Group')

也出现了相同的FieldError错误。

奇怪的是，这个方法却可以正常工作：

try:
    Group.objects.get(name="Active Rater")    #crazily not working
except django.core.exceptions.FieldError as e:
    group = Group.objects.get(name="Active Rater")    #crazily works

我还没有深入研究Django的代码，搞清楚为什么会这样。