I like the is_correct and the count attributes!
The problem with that it is hard to make a quiz in the admin section. You
create a quiz and a question but then you have to go to the choice section
and create choices for the question. And even if you have the same choice
in two questions you need to create the choice multiple times.
It would be nice to have text fields and be able to say add a 5th choice,
and just enter text. It seems easier than changing forms and selecting an
option from a large list, or even in your case creating each choice for a
quiz.
Is there a way to have the admin link from Question to Choice in this
configuration?
On Wednesday, 10 April 2013 14:43:54 UTC-4, BFSchott wrote:
>
> Is a many-to-many what you really want? How common are the choices and
> suppose someone goes in an edits one of the choices and suddenly it doesn't
> match up right with a different linked question. It might make sense to
> use a reverse foreign key one-to-many relationship and provide a mechanism
> to duplicate a question, including its member choices. That way you could
> add some statistics on each choice unique to a question.
>
> Several possible paths depending on how you want to go.
>
> 1. You could just use a reverse foreign key and have a one-to-many
> relationship.
>
> class Choice(models.Model):
> question = models.ForeignKey('Question', related_name='choices')
> choice = models.CharField(max_length=255)
> index = models.IntegerField(default=0) # in case you want order_by
> filtering
> count = models.IntegerField(default=0) # in case you want to count number
> of times this is selected right or wrong
> is_correct = models.BooleanField(default=False) # boolean flag replaces
> "answer"
>
> class Question(models.Model):
> question = models.CharField(max_length=64)
> module = ....
>
> def times_correct(self):
> return self.choices.objects.filter(is_correct=True).aggregate(Sum('count'))
>
> def times_total(self):
> return self.choices.objects.aggregate(Sum('count'))
>
>
> Then in the admin, you create a table inline admin and just add unique
> entries.
>
>
> 2. Encode the thing as a JSON list and store in a JSONField
> https://github.com/derek-schaefer/django-json-field
>
> 3. There are several flavors of ListField out there:
> http://djangosnippets.org/snippets/1491/
>
>
>
> Brian Schott
> bfsc...@gmail.com <javascript:>
>
>
>
> On Apr 10, 2013, at 1:38 PM, Cody Scott <cody.j....@gmail.com<javascript:>>
> wrote:
>
> I am trying to store questions in a database.
> I don't to set a fixed number of options for the question, some questions
> could have 4 or 2 or 5.
>
> Currently I am using a ManyToManyField to a table that just contains a
> CharField.
> This works but creating an option requires making another Choice object
> and selecting that, also when you want to select a Choice that has already
> been created you have to use that little box in the admin and it doesn't
> scale when you have hundreds of options.
>
> Even if I wanted to have 4 options every time what is the recommended way
> without having four CharFields?
>
> class Choice(models.Model):
> choice = models.CharField(max_length=255)
> def __unicode__(self):
> return self.choice
> #multiple choice questionclass Question(models.Model):
> question = models.CharField(max_length=64)
> answer = models.CharField(max_length=255)
> choices = models.ManyToManyField(Choice, related_name='questions',
> verbose_name='options')
> module = models.ForeignKey('Module', related_name='questions')
>
> times_correct = models.IntegerField(editable=False, default=0)
> times_total = models.IntegerField(editable=False, default=0)
>
> def _get_average(self):
> "Returns the average in percent"
> if self.times_total != 0:
> return (self.times_correct / float(self.times_total))*100
> return 0.0
> average = property(_get_average)
>
> def __unicode__(self):
> return self.question
>
>
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