django object has no attribute status_code

Question

I am learning to develop and trying to develop a screen in django 1.1 and I am getting this error below. I already took a look at some stacks I already looked at httpresponse, however, I was not successful could someone help me and explain what could be wrong?

I'm getting this error on the console:

Internal Server Error: /gerencia-margem/list
Traceback (most recent call last):
File "/home/murilo/virtualenv/intranet_erp/local/lib/python2.7/site-packages/django/core/handlers/base.py", line 131, in get_response
response = middleware_method(request, response)
File "/home/murilo/virtualenv/intranet_erp/local/lib/python2.7/site-packages/django/middleware/locale.py", line 38, in process_response
if (response.status_code == 404 and not language_from_path and
AttributeError: 'HistoricoComissao' object has no attribute 'status_code'
[18/Nov/2020 13:30:06] "GET /gerencia-margem/list HTTP/1.1" 500 77145

This is models,

# -*- coding: utf-8 -*-
from django.db import models


class HistoricoComissao(models.Model):
class Meta:
    verbose_name = u'Historico Comissão'
    verbose_name_plural = u'Historico Comissões'

pednum = models.CharField(max_length=40, blank=True)
margem = models.DecimalField(decimal_places=4, max_digits=15, null=True, blank=True) 
data = models.DateTimeField()
historico = models.TextField((u'Historico HTML'), blank=True)

status = models.PositiveSmallIntegerField(choices=[(1, 'Em Aberto'),
                                                   (3, 'Pendente'),
                                                   (4, 'Finalizado'),])

def __unicode__(self):
        return str(self.pednum)

this is the views

from django.views import View
from django.shortcuts import render
from django.http import HttpResponseRedirect, HttpResponse
from django.core.urlresolvers import reverse
from ..models import HistoricoComissao


def listview(request):
template_name = 'comissao/margenslist.html'
comissoes = HistoricoComissao.objects.all
context = {
    'comissoes': comissoes
}
return render(request,template_name,context)

this is urls.py

from django.conf.urls import  url
from . import views


urlpatterns = [
    url(r'^list$', views.HistoricoComissao, name='historico_comissao'),
 ]

willeM_ Van Onsem · Accepted Answer · 2020-11-19 00:12:32Z

3

In your urls.py you use as view HistoricoComissao, but that is a re-export of a Django model, not of a view. You should use the listview:

from django.conf.urls import  url
from . import views


urlpatterns = [
    # link to a view function ↓
    url(r'^list$', views.listview, name='historico_comissao'),
]

I am learning to develop and trying to develop a screen in django 1.1.

Please upgrade. Django-1.1 was end of life before 2013. Furthermore a lot of things have changed in the meantime (and often improved).

answered Nov 19, 2020 at 0:12

willeM_ Van Onsem

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1 Comment

jefferson lopes da silva Over a year ago

Thanks for helping me, I'm studying Django 1.1, because some of the company's projects are in this model, so I'm starting with it.começando com ele.

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