Question

I know this is a very basic concept in Django, and I have tried the tutorial but it is not working. I am working on a comic book database with the models set like this (at least, a sample of two of the models):

Class Title(models.Model):
    title = models.CharField(max_length=256)
    vol = models.IntegerField("Vol.")
    slug = models.SlugField(blank=True, null=True) 
    #desc = models.CharField(max_length=256)   
    class Meta:
        ordering = ['title']
    def get_absolute_url(self):
        return "/comics2/title/%s" % self.slug        
    def __unicode__(self):
        return self.title


class Issue(models.Model):
    title = models.ForeignKey(Title)
    number = models.IntegerField(help_text="Enter the number only. Do not include the hashtag.")
    writer = models.ManyToManyField(Creator)

What I am trying to do is create a page that shows a list of all the issues within that Title.

But, I have it setup in the views like this:

class AstonishingXMenIssueListView(ListView):

    context_object_name = "astonishing_list"
    queryset = Issue.objects.filter(title__title="Astonishing X-Men").order_by("number")
    template_name = "comics2/astonishing_list.html"

My urls.py look like this:

(r'^comics2/title/(?P<title_slug>[-\w]+)/$', AstonishingXMenIssueListView.as_view(
    )), 

Of course, going to /uncanny-xmen-v1/ shows the same thing as the Astonishing link above.

Obviously, this is not a practical way to list issues by title for future issues and titles, so I need it setup so that I don’t have to individually do this. Now, I have tried following the Django generic views tutorial, but I got an index tuple error.

I’ve tried this, but it doesn’t work. This is what gets me the index tuple error.

class IssuesByTitleView(ListView):

    context_object_name = "issues_by_title_list"
    template_name = "comics2/issues_by_title.html",

    def get_queryset(self):
        title = get_object_or_404(Title, title__iexact=self.args[0])
        return Issue.objects.filter(title=title)

Any ideas? And can someone please reply in baby-language, as I am new to Django and Python, so simply telling me to look at the Tutorial again isn’t going to help. So, maybe writing out the code would help! Thanks!

Answer 1

Generally, your IssueByTitleView is the right way to do it. But as you use named groups in your URL regex (the (?P<title_slug>[-\w]+) part of your URL), you have to access the URL parameters through self.kwargs instead of self.args. Also, you have to filter on the slug field, not the title field:

title = get_object_or_404(Title, slug=self.kwargs['title_slug'])