Djangopython:NoReverseMatch问题：我应该给它什么参数？

Question

我有以下看法：

def default_detail (request, equipment_id):
    equipment = Equipment.objects.get(id = equipment_id)
    context = {'equipment' : equipment}
    return render(request, 'calbase/default_detail.html', context)


def default_add_cal (request, equipment_id):
    equipment = get_object_or_404(Equipment, id = equipment_id)
    EquipmentInlineFormSet = inlineformset_factory(Equipment, Calibration, fields = ('cal_by', 'cal_date', 'notes'), extra = 1, )
    if request.method == "POST":
        if 'calibration' in request.POST:
            formset = EquipmentInlineFormSet(request.POST, request.FILES, instance=equipment)
            if formset.is_valid():
                return HttpResponseRedirect(reverse('calbase:default_detail', args=(equipment_id)))
    else:
        formset = EquipmentInlineFormSet(instance=equipment)
    return render(request, 'calbase/default_detail_cal.html', {'formset' : formset, 'equipment' : equipment})

URL如下所示：

urlpatterns = [
    url(r'^$', views.default, name = 'default'),
    url(r'^default/(?P<equipment_id>[0-9]+)/$', views.default_detail, name = 'default_detail'),
    url(r'^default/new/$', views.default_new, name = 'default_new'),
    url(r'^default/(?P<equipment_id>[0-9]+)/cal/$', views.default_add_cal, name = 'default_add_cal'),
]

default\u add\u cal、default\u detail\u cal.html的模板如下：

{% block content %}
<form method="POST" action = "{% url 'calbase:default_add_cal' equipment.id %}">{% csrf_token %}
        {{ formset }}
        <button type="submit" class="save btn btn-default" name = "calibration">Save</button>
     </form>

<a href="{% url 'calbase:default' %}">Back?</a>
{% endblock %}

所以校准是设备的外键：一个设备可以有多个校准，这里我尝试使用inlineformset让用户向设备添加校准。问题是我得到了这个：

NoReverseMatch at /calbase/default/41/cal/ Reverse for 'default_detail' with arguments '('4', '1')' and keyword arguments '{}' not found. 1 pattern(s) tried: ['calbase/default/(?P[0-9]+)/$']

例如，对于设备id=41，当我这样做时

return HttpResponseRedirect(reverse('calbase:default_detail', args=(equipment_id)))

实际传递的参数是“（'4'，'1'）”而不是41。这怎么可能呢？我该怎么解决这个问题？谢谢

Answer 1

应该将参数作为元组传递。这将使args成为一个iterable，equipment_id是其中一个参数：

args=(equipment_id,‌​)
#                 ^

这些分组括号不会使参数成为元组：

>>> args = (42)
>>> type(args)
<class 'int'>
>>>
>>> args = (42,)
>>> type(args)
<class 'tuple'>

即使没有父母也能工作：

>>> args = 42,
>>> type(args)
<class 'tuple'>