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PODjango QuerySets: Get unique/first element from ForeignKey field
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copied!<p>I've following models:</p> <pre><code>class SchoolClass(models.Model): title = models.CharField() max_students = models.IntegerField(default=10) is_published = models.BooleanField(default=False) class Session(models.Model): school_class = models.ForeignKey(SchoolClass, related_name="sessions") start_date = models.DateField() enrolled_students = models.IntegerField() </code></pre> <p>A SchoolClass can have multiple Sessions (e.g. "Maths for Dummies" on Monday, Tuesday and Friday = 3 Sessions).</p> <p>On my Index page (ListView), I'd like to show all upcoming classes (which have at least one session) only once, since classes which have a lot of sessions would make the overview messy.</p> <p>How can I create a list of Session-Objects (I can get the relevant course info via session.school_class), which comply with following requirements:</p> <ul> <li><code>start_date__gte=datetime.now().date()</code></li> <li><code>school_class.is_published=True</code></li> <li>it's the first elelement of SchoolClass (<code>oder_by('start_date')[0]</code>)</li> </ul> <p>Thanks!</p> <p>PS: I'm running MySQL as DB, so distinct('field') won't work..</p> <hr> <p><strong>Edit:</strong></p> <p>Below is my current solution, which works, but is not very clean (redundant hits on the DB)</p> <pre><code>def get_sessions_mixin(filter_args): session_qs = [] qs = SchoolClass.objects.filter(**filter_args) for schoolclass in qs: session_qs.append(schoolclass.get_first_session()) #session_qs = Session.objects.filter(course_meta__is_deleted=False, date_from__gte=datetime.now().date()).\ # annotate(Max('date_from')) #session_qs = qs.order_by('get_first_session__date_from') return session_qs </code></pre> <p>and inside models.py</p> <pre><code>class Sessions(models.Model): # ... def get_first_session(self): if self.is_published: tmp_sessions = self.sessions.filter(date_from__gte=datetime.now().date()) if tmp_sessions: return tmp_sessions.order_by('date_from')[0] return None </code></pre>

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