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copied!<p>While adam's solution is correct, a more django-ish way would be to use a manager.</p> <p>See <a href="http://www.b-list.org/weblog/2008/feb/25/managers/" rel="nofollow noreferrer">Managers versus class methods</a> on James Bennett's blog</p> <p>Among other advantages :</p> <ul> <li>a manager carries all query-specific code, while avoiding to clutter the model class</li> <li>the same manager class can be shared among several classes</li> <li>the manager can be used directly on a model class, or via a one-to-many or m2m relation</li> </ul> <p>Thus, for the above question :</p> <pre><code>from django.db import models from django.db.models.manager import Manager class ProductScoreManager(Manager): use_for_related_field = True def top_score(self, limit=2): return self.get_query_set().order_by('-score')[:limit] </code></pre> <p>Then add this manager class as <em>default manager</em> for Product :</p> <pre><code>class Product(models.Model): ... objects = ProductScoreManager() ... </code></pre> <p>Now, thanks to <em>objects</em> overriding the default manager, and <em>use_for_related_field</em> allowing its use in relation queries, the top_score method can be used in any model related to products.</p> <pre><code>myproducttype = ProductType.objects.get(...) myproducttype.products.top_score() # return top 2 products </code></pre> <p>This allows a more consistent syntax : the related products is always accessed via <em>products</em>, <em>top_score</em> acting as a filter. Additionally, ProductType class is no more cluttered with Product's query logic.</p>

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