Neue App, die AttributeError verursacht: 'str' -Objekt hat kein Attribut '_meta' in Python/Django App

Question

Mit Python 2.7 und Django 1.9.9 Ich erhalte den folgenden Fehler, wenn ich eine App einschließe, die ich gerade entwickle mein INSTALLED_APS

Traceback (most recent call last): 
    File "manage.py", line 22, in <module> 
    execute_from_command_line(sys.argv) 
    File "/var/www/cltc/env/local/lib/python2.7/site-packages/django/core/management/__init__.py", line 353, in execute_from_command_line 
    utility.execute() 
    File "/var/www/cltc/env/local/lib/python2.7/site-packages/django/core/management/__init__.py", line 345, in execute 
    self.fetch_command(subcommand).run_from_argv(self.argv) 
    File "/var/www/cltc/env/local/lib/python2.7/site-packages/django/core/management/base.py", line 348, in run_from_argv 
    self.execute(*args, **cmd_options) 
    File "/var/www/cltc/env/local/lib/python2.7/site-packages/django/core/management/base.py", line 398, in execute 
    self.check() 
    File "/var/www/cltc/env/local/lib/python2.7/site-packages/django/core/management/base.py", line 426, in check 
    include_deployment_checks=include_deployment_checks, 
    File "/var/www/cltc/env/local/lib/python2.7/site-packages/django/core/checks/registry.py", line 75, in run_checks 
    new_errors = check(app_configs=app_configs) 
    File "/var/www/cltc/env/local/lib/python2.7/site-packages/django/core/checks/model_checks.py", line 28, in check_all_models 
    errors.extend(model.check(**kwargs)) 
    File "/var/www/cltc/env/local/lib/python2.7/site-packages/django/db/models/base.py", line 1180, in check 
    errors.extend(cls._check_long_column_names()) 
    File "/var/www/cltc/env/local/lib/python2.7/site-packages/django/db/models/base.py", line 1631, in _check_long_column_names 
    for m2m in f.remote_field.through._meta.local_fields: 
AttributeError: 'str' object has no attribute '_meta' 

Das ist, glaube ich durch etwas verursacht wird falsch in models.py, die wie folgt aussieht:

from django.db import models 
from django.contrib.auth.models import User 
from django.conf import settings 
import datetime 

from django.core.exceptions import ValidationError 
from django.utils.translation import ugettext_lazy as _ 

class Category(models.Model): 
    name = models.CharField('Category', max_length=30) 
    age = models.IntegerField('Member age at start of Subscription', default=18) 

class Subscription(models.Model): 
    name = models.CharField('Subscription', max_length=30) 
    cost = models.DecimalField('Price', max_digits=6, decimal_places=2, default=0.00) 
    start = models.DateField('Start Date') 
    end = models.DateField('End Date') 
    category = models.ManyToManyField(
     Category, 
     through   = 'SubscritptionCategory', 
     related_name = 'category', 
     verbose_name = 'Membership Category', 
     help_text  = 'Membership Categories included in the Subscription' 
     ) 

    def __unicode__(self): 
     return u'%s' % (self.name) 

    def clean (self): 
     if self.start > self.end: 
      raise ValidationError(
       _("Start Date must be earlier than End Date"), 
       ) 

    def is_live(self): 
     if self.end >= datetime.datetime.now().date(): 
      return True 
     else: 
      return False 

class SubscriptionCategory (models.Model): 
    subscription = models.ForeignKey(
     Subscription, 
     verbose_name = 'Subscription', 
     help_text  = 'A class of membership (which could include several members, eg Family).', 
    ) 
    category = models.ForeignKey(
     Category, 
     verbose_name = 'Category', 
     help_text  = 'A class of member (eg adult)', 
    ) 

Jede Hilfe sehr willkommen

Answer 1

Sie haben einen Tippfehler in yo ur Erklärung des through Attributs von Subscription.category: "SubscriptionsCategory" statt "SubscriptionCategory". Aus diesem Grund kann Django das Modell, auf das Sie verweisen, nicht finden.

Beachten Sie jedoch, dass, da Sie keine zusätzlichen Felder auf diesem Modell definieren, es nicht viel Sinn hat; Ihr Code wäre einfacher und viele von Djangos Funktionen würden besser funktionieren, wenn Sie sie nicht definieren würden.