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POHow can I force django to use the lowest model instance?
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copied!<p>Suppose I have my models set up like this:</p> <pre><code>class B(Model): ... def __str__(self): return "B" class C(B): ... def __str__(self): return "C" class A(Model): b = ForeignKey(B) def __str__(self): print "A: %s"%(self.b) a = A(b=C(...)) a.save() print str(a) # Prints "A: B" </code></pre> <p>Maybe I'm a bit confused how django inheritance works. My intention is to have the program print <code>"A: C"</code> at runtime (since <code>A.b</code> is an instance of model <code>C</code>)</p> <p>I think I <em>might</em> be asking about <a href="https://docs.djangoproject.com/en/dev/topics/db/models/#multi-table-inheritance" rel="nofollow">multi-table inheritance</a>. But that's only if you know what instance of a subclass you want.</p> <p>As another example of how I'm confused, I'd like to borrow the example from the docs:</p> <pre><code># Assume Restaurant, Park, and Museum inherit Place bills = Restaurant.objects.create(name="Bill's Pub", burger="Bill's Burger") city_prk = Park.objects.create(name='City Park', num_trails=5) nose = Museum.objects.create(name='Nose Museum', est=1940) places = Places.objects.all() </code></pre> <p>I'll definitely get a list of 3 objects (of type <code>Place</code>) but I have no way of differentiating the types of places. For example, if I want to print out the str value of each place...</p> <pre><code>for place in places: print str(place) </code></pre> <p>...python will just execute <code>Place.__str__()</code> and not <code>Restaurant.__str__()</code> (for the restaurant) or <code>Park.__str__()</code> (for the park). This doesn't seem to happen with "normal" python. Python should normally automatically find the lowest-inherited class and execute the overridden function from that (if B overrides a method from A, B's method will be executed).</p> <p>Are my assumptions correct? Sorry for the long question (and sorry if it's not clear). I'm not exactly sure how to ask what I'm wondering.</p>

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