templates - Want to print out a list of items from another view django -

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i have view displays list items.

def edit_order(request, order_no):  try:     status_list = models.status.objects.all()     order = models.order.objects.get(pk = order_no)     if order.is_storage:         items = models.storageitem.objects.filter(orderstoragelist__order__pk = order.pk)     else:         items = models.storageitem.objects.filter(orderservicelist__order__pk = order.pk) except:     return httpresponsenotfound()

i want put these list of item in view. unfortunately proving trickier thought.

@login_required def client_items(request, client_id = 0):     client = none     items = none     try:         client = models.client.objects.get(pk = client_id)         items = client.storageitem_set.all()         item_list = models.storageitem.objects.filter(orderstoragelist__order__pk = order.pk)     except:         return httpresponse(reverse(return_clients))     return render_to_response('items.html', {'items':items, 'client':client, 'item_list':item_list}, context_instance = requestcontext(request))

i thought maybe can paste definition of items , call item_list not work. ideas

items.html

{% item in item_list %}     {{item.tiptop_id} {% endfor %}

from comment:

i white screen url printed on screen. /tiptop/client in case.

because that's you've asked for:

except:     return httpresponse(reverse(return_clients))

this means if there bugs or problems in above, view output response containing url. maybe meant use httpresponseredirect, browser redirects url - still should not use blank except, prevents seeing going wrong.

to answer main question, think edit_order view returns: gives complete html response rendered template. how use element in query in view? need think logically this.

one possible solution define separate function returns data want - plain queryset - , both views can call it. want?


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