遵循Django实用教程得到这个错误。我已经搜索过这里和谷歌,但没有找到任何关于这一点的信息,所以我想这将是我很容易错过的东西。甘比诺例外_标签.py公司名称:
context[self.varname] = self.model._default_manager.all()[:self.num]
在模型.py在
^{pr2}$甘比诺_标签.py在
from django import template
from django.db.models import get_model
def do_latest_content(parser, token):
bits = token.contents.split()
# Checks to make sure that there are 5 words in the tag
if len(bits) != 5:
raise template.TemplateSyntaxError("'get_latest_content' tag takes exactly four arguments")
model_args = bits[1].split('.')
# Checks to make sure first argument is appname.model name
if len(model_args) != 2:
raise template.TemplateSyntaxError("First argument to 'get_latest_content' must be an 'application name'.'model name' string")
model = get_model(*model_args)
# Checks to make sure model is != none
if model is None:
raise template.TemplateSyntaxError("'get_latest_content' tag got an invalid model: %s" % bits[1])
return LatestContentNode(bits[1], bits[2], bits[4])
class LatestContentNode(template.Node):
def __init__(self, model, num, varname):
self.model = model
# convert num to int
self.num = int(num)
self.varname = varname
def render(self, context):
# uses default manager for Entry.live.all() instances
context[self.varname] = self.model._default_manager.all()[:self.num]
return ''
register = template.Library()
register.tag('get_latest_content', do_latest_content)
在基本.html在
{% load gambino_tags %}
<h2>Five latest entries:</h2>
<ul>
{% get_latest_content gambino.entry 5 as latest_entries %}
{% for entry in latest_entries %}
<li>
<a class="list" href="{{ entry.get_absolute_url }}">{{ entry.title }}</a>
Posted {{ entry.pub_date|timesince }} ago.
</li>
{% endfor %}
</ul>
<h2>Five latest links:</h2>
<ul>
{% get_latest_content gambino.link 5 as latest_links %}
{% for link in latest_links %}
<li>
<a href="{{ link.get_absolute_url }}">{{ link.title }}</a>
Posted {{ link.pub_date|timesince }} ago.
</li>
{% endfor %}
</ul>
谢谢!在
您的
Node
的model
参数是一个字符串(或者至少,这就是您要传递的内容),而不是Model
对象。在也许你真的是说。。。在
^{pr2}$是吗?在
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