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PODjango - Pass model name as parameter to generic view
 

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  1. PODjango - Pass model name as parameter to generic view
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    <p>Let's say I have some Models that inherit from a base class, Animal. I can use generic views and route Cat/12 to a detail view and Dod/10 to the same detail view with a different context. But I would like to get the Model name from the url so that I dont have to define a route.</p> <p>I have something like this:</p> <pre><code>url(r'^cat/(?P&lt;slug&gt;[-\w]+)/$', DetailView.as_view( queryset=Cat.objects.filter(), model=Cat, context_object_name='animal', template_name='animal/detail.html'), name='detail'), url(r'^dog/(?P&lt;slug&gt;[-\w]+)/$', DetailView.as_view( queryset=Dog.objects.filter(), model=Dog, context_object_name='animal', template_name='animal/detail.html'), name='detail'), ... </code></pre> <p>Obviously, this is too much repetitive code. I would rather do something like this:</p> <pre><code>url(r'^?P&lt;my_animal&gt;\w+/(?P&lt;slug&gt;[-\w]+)/$', DetailView.as_view( queryset=my_animal.objects.filter(), model=my_animal, context_object_name='animal', template_name='animal/detail.html'), name='detail'), ... </code></pre> <p>Can I do this?</p> <p><strong>EDIT</strong></p> <p>Here's what I ended up with, thanks to Darwin's help. It avoids the if/else to get the Model name:</p> <pre><code>class AnimalDetailView(DetailView): context_object_name='animal' template_name='animals/detail.html' def dispatch(self, request, *args, **kwargs): my_animal = kwargs.get('my_animal', None) self.model = get_model('animals',my_animal.capitalize()) try: ret = super(AnimalDetailView, self).dispatch(request, *args, **kwargs) except AttributeError: raise Http404 return ret def get_queryset(self): return self.model.objects.filter() </code></pre> <p>Next time I have a question about Inheritance, I'll consult Darwin! LoL</p>
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