AttributeError at / 'str' object has no attribute 'objects'

Question

I am new to Django. I am working on a project that uses weather API to get the weather.Everything was working fine until models.py was made and imported city on views.py

I use ver. 1.11.13

models.py

  from __future__ import unicode_literals

  from django.db import models

  class city(models.Model):
  name = models.CharField(max_length=25)

  def __str__(self):
      return self.name

  class Meta:
        verbose_name_plural ='cities'

views.py ( Error comes at cities = city.objects.all())

import requests

from django.shortcuts import render
from .models import city

def index(request):
  url= 'http://api.openweathermap.org/data/2.5/weather?q={}& units=imperial&appid=e1a2ef38103d2e572d316e38452e2acd'
  city = 'Lucknow'

  cities = city.objects.all()

  weather_data =[]

  for city in cities:


    r= requests.get(url.format(city)).json()


    city_weather = {
         'city':city.name ,
         'temperature' :r['main']['temp'],
         'description' :r['weather'][0]['description'],
         'icon' :r['weather'][0]['icon'] ,
    }
    weather_data.append(city_weather)

    print(weather_data)

context = {'weather_data' : city_weather}

return render(request,'weather/weather.html', context)

You have overwritten city with city = 'Lucknow'. So rename the variable. Furthermore Python style guidlines say to let classes start with an uppercase — willeM_ Van Onsem
– willeM_ Van Onsem, Commented Jun 13, 2018 at 22:26

willeM_ Van Onsem · Accepted Answer · 2018-06-13 22:29:29Z

3

In your index function, you write:

def index(request):
    url= 'http://api.openweathermap.org/data/2.5/weather?q={}& units=imperial&appid=e1a2ef38103d2e572d316e38452e2acd'
    city = 'Lucknow'

    cities = city.objects.all()
    # ...

As a result, you defined city here as a locally scoped variable, and with a string value. A quick fix is to rename the string, like:

def index(request):
    url= 'http://api.openweathermap.org/data/2.5/weather?q={}& units=imperial&appid=e1a2ef38103d2e572d316e38452e2acd'
    city_ = 'Lucknow'

    cities = city.objects.all()
    # ...
    # use city_ for the string element

But nevertheless, a class typically starts with an uppercase, so I would advice to rename your class City instead of city.

Furthermore I find it weird that you defined this string, since nowehere in the view, you are supposed to use it. The url.format(..) should probably use the city.name of the City object in the queryset.

answered Jun 13, 2018 at 22:29

willeM_ Van Onsem

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7 Comments

Adarsh Bahadur Over a year ago

Thank you that Helped !! another quick help why does the value prints like 'city': u'New Delhi', 'icon': u'50n', 'temperature': 98.6, 'description': u'dust'}] why does that u appears before the values ?

willeM_ Van Onsem Over a year ago

@AdarshBahadur: you probably print {{weather_data}}. Then it prints a representation of the entire dictionary. You should use {{ weather_data.city }}, {{ weather_data.description }}, to access the individual members.

Adarsh Bahadur Over a year ago

Okay. Thank you for your time !

Adarsh Bahadur Over a year ago

Completely removing this city = 'Lucknow' helped. I used this to test my app initially.

willeM_ Van Onsem Over a year ago

@AdarshBahadur: ah, then the issue is now resolved. Although I would advice to rename your classes such that these start with an uppercase. It is recommended by PEP-8: python.org/dev/peps/pep-0008

|

Nids Barthwal · Accepted Answer · 2020-07-05 02:27:31Z

3

Solve this issue.

For dynamic table name you can use like

 table_name = 'MyTable'
 model = getattr(YOURAPP.models, table_name)
 model.objects.all()

answered Jul 5, 2020 at 2:27

Nids Barthwal

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