使用模板标记在html模板中求和

sus*_*ush 2 python django django-templates django-template-filters

我想在HTML中求和,但模板标签返回0,

View.py

def gen_Report(request):

### query returns below output 
list=[{'total': 1744, 'user': u'x'}, {'total': 13, 'user': u'y'}, {'total': 126, 'user': u'z'}, {'total': 46, 'user': u'm'}, {'total': 4, 'user': u'n'}, {'total': 8, 'user': u'o'},  {'total': 3, 'user': u'p'}]

return render_to_response('user.html', locals(),
                            context_instance = RequestContext(request))
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模板:

user.html

  {% load temptags %}

 <table id="myTable" class="tablesorter">
    <thead>
    <tr>

    <th>S.No</th>
    <th>role</th>
    <th>Count</th>

    </tr>
    </thead>
    {% for fetch in list %}

    <tr>
    <td>{{forloop.counter}}</td>
    <td>{{fetch.user}}</td>
    <td>{{fetch.total}}</td>



    {% endfor %}
    <td>{{ list.total|running_total}}</td>
    <tr>

    </table>
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模板标签:

from django.template import Library
register = Library()
@register.filter
def running_total(list_total):
  return sum(d.get('list_sum') for d in list_total)
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输出:

S.No    user          Count
1     x       1744
2     y         13
3     z         126
4     m         46
5     n              4
6     o          8
Sum------------------>   0  (it returns zero)
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我在这里做错了什么?

你可以帮助我,如何在这里使用模板标签返回总和?

dr *_*bob 7

您的模板标记看起来不对.你有role_total参数,然后迭代list_total(看似未定义),并从列表中的每个字典尝试获取list_sum看似未定义的键.

from django.template import Library
register = Library()
@register.filter
def running_total(your_dict_list):
   return sum(d['total'] for d in your_dict_list)
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并在as模板中调用它 <td>{{ list|running_total}}</td>