Question

我从db获得了2个查询集：

all_locations = Locations.objects.all()[:5]
rating = Rating.objects.all()[:5]
return render_to_response('index.html',{'all':all_locations,'rating':rating},context_instance=RequestContext(request))

但我被困在这里，不知道如何在一个循环中循环这两个查询集。这是错误的：

{% if all and rating %}
  {% for every in all and rating  %}
         {{every.locationname}}, {{every.rating_score}}
  {% endfor %}
{% endif %}

Answer 1

您可以尝试zip(all_locations, rating)。它将生成一个元组列表。然后你可以成对迭代它们。这是一个例子：（demo）

all_locations = ['ca','ny','fl']
ratings = ['best','great','good']
for (l,r) in zip(all_locations,ratings): 
   print l+':'+r

输出

ca:best
ny:great
fl:good

Answer 2

我也遇到了这个问题。现在，我已修复它。我正在使用

new=tuple(zip(queryset1,queryset2))  
return render(request, 'template.html', {"n": new}).

在view.py中。

在template.html中，我使用以下三个句子。

{% for i in n %}

{% for j in i|slice:"0:1" %}

......operate queryset1

{% endfor %}
{% for z in i|slice:"1:2" %}

.....operate queryset2


{% endfor %}

{% endfor %}

It seems this method will fulfill your needs.

Answer 3

这可能有效：

{% with rating|length as range %}
    {% for _ in range %}
        {{ rating[forloop.counter] }}
        {{ location[forloop.counter] }}
    {% endfor %}
{% endwith %}

我不确定rating|length是否会胜任该工作......您可能需要添加rating|length|times' with次`过滤器定义为：

@register.filter(name='times') 
def times(number):
    return range(number)

django模板 - 用于循环超过2个查询集

3 个答案: