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  1. POusing django, how to create method in model to return query data
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    <p>In the following code, I'm tyring to create the method show_pro that will show all Arguments for the Case that are pro. </p> <p>I am getting this error:</p> <pre><code>&gt;&gt;&gt; Case.objects.all()[0].show_pro() Traceback (most recent call last): File "&lt;console&gt;", line 1, in &lt;module&gt; File "/home/john/mysite/../mysite/cases/models.py", line 23, in show_pro return self.objects.filter(argument__side__contains='p') File "/usr/lib/python2.5/site-packages/django/db/models/manager.py", line 151, in __get__ raise AttributeError, "Manager isn't accessible via %s instances" % type.__name__ AttributeError: Manager isn't accessible via Case instances </code></pre> <p>here is the code: </p> <pre><code>from django.db import models from django.contrib.auth.models import User import datetime SIDE_CHOICES = ( ('p', 'pro'), ('c', 'con'), ('u', 'undecided'), ) class Case(models.Model): question = models.CharField(max_length=200) owner = models.ForeignKey(User) pub_date = models.DateTimeField('date published') rating = models.IntegerField() def __unicode__(self): return self.question def was_published_today(self): return self.put_date.date() == datetime.date.today() def show_pro(self): return self.objects.filter(argument__side__contains='p') class Argument(models.Model): case = models.ForeignKey(Case) reason = models.CharField(max_length=200) rating = models.IntegerField() owner = models.ForeignKey(User) side = models.CharField(max_length=1, choices=SIDE_CHOICES) def __unicode__(self): return self.reason </code></pre>
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