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我正在尝试编写一个简单的算法来获得 django mptt 模型的“等级”。“等级”是指后代的级别数......类似于 MPTTModel.get_level() 但从叶到根计数。

我的方法如下：

视图.py

 def index(request):
    nodes = Node.objects.all()
    nodesets = []
    for i in nodes:
        figure_rank(i, i)
        if i.count >= 3:
            nodeset = Node_set(node=i, name=i.name)
            nodeset.save()      # This is where the error is caught
            nodesets.append(dataset)
    return render(request, 'associate/index.html', {'nodesets':nodesets})

def figure_rank(over, under):

    if under.get_children()[0]:
        r = under.get_children()[0]
        over.count += 1
        figure_rank(over, r)
    else:
        return over.count

对于 rank >=3 的节点，我想创建一个 Node_set 并在它们都给定等级后返回一组节点。

我收到以下错误：

Environment:


Request Method: GET
Request URL: http://127.0.0.1:8000/associate/

Django Version: 1.6.2
Python Version: 2.7.5
Installed Applications:
('django.contrib.admin',
 'django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.messages',
 'django.contrib.staticfiles',
 'associate',
 'images_app',
 'mptt')
Installed Middleware:
('django.contrib.sessions.middleware.SessionMiddleware',
 'django.middleware.common.CommonMiddleware',
 'django.middleware.csrf.CsrfViewMiddleware',
 'django.contrib.auth.middleware.AuthenticationMiddleware',
 'django.contrib.messages.middleware.MessageMiddleware',
 'django.middleware.clickjacking.XFrameOptionsMiddleware')


Traceback:
File "C:\Python27\lib\site-packages\django\core\handlers\base.py" in get_response
  114.                     response = wrapped_callback(request, *callback_args, **callback_kwargs)
File "C:\djangoprojects\images\associate\views.py" in index
  9.         figure_rank(i, i)
File "C:\djangoprojects\images\associate\views.py" in figure_rank
  23.         figure_rank(over, r)
File "C:\djangoprojects\images\associate\views.py" in figure_rank
  23.         figure_rank(over, r)
File "C:\djangoprojects\images\associate\views.py" in figure_rank
  20.     if under.get_children()[0]:
File "C:\Python27\lib\site-packages\django\db\models\query.py" in __getitem__
  132.         return list(qs)[0]

Exception Type: IndexError at /associate/
Exception Value: list index out of range

我在这里做错了什么？

4

1 回答 1

1

您需要正确检查是否没有孩子。在figure_rank,under.get_children()返回一个空列表，然后您尝试访问第一个元素。首先检查列表是否为空。

def figure_rank(over, under):
    children = under.get_children()
    if chilren: # check if children is nonempty before accessing the first element
        r = children[0]
        over.count += 1
        figure_rank(over, r)
    else:
        return over.count
于 2014-05-12T19:06:48.720 回答