Union and Intersect in Django

hamdiakoguz picture hamdiakoguz · Sep 20, 2008 · Viewed 15.3k times · Source
class Tag(models.Model):
  name = models.CharField(maxlength=100)

class Blog(models.Model):
  name = models.CharField(maxlength=100)
  tags =  models.ManyToManyField(Tag)

Simple models just to ask my question.

I wonder how can i query blogs using tags in two different ways.

  • Blog entries that are tagged with "tag1" or "tag2": Blog.objects.filter(tags_in=[1,2]).distinct()
  • Blog objects that are tagged with "tag1" and "tag2" : ?
  • Blog objects that are tagged with exactly "tag1" and "tag2" and nothing else : ??

Tag and Blog is just used for an example.

Answer

Clint Ecker picture Clint Ecker · Sep 20, 2008

You could use Q objects for #1:

# Blogs who have either hockey or django tags.
from django.db.models import Q
Blog.objects.filter(
    Q(tags__name__iexact='hockey') | Q(tags__name__iexact='django')
)

Unions and intersections, I believe, are a bit outside the scope of the Django ORM, but its possible to to these. The following examples are from a Django application called called django-tagging that provides the functionality. Line 346 of models.py:

For part two, you're looking for a union of two queries, basically

def get_union_by_model(self, queryset_or_model, tags):
    """
    Create a ``QuerySet`` containing instances of the specified
    model associated with *any* of the given list of tags.
    """
    tags = get_tag_list(tags)
    tag_count = len(tags)
    queryset, model = get_queryset_and_model(queryset_or_model)

    if not tag_count:
        return model._default_manager.none()

    model_table = qn(model._meta.db_table)
    # This query selects the ids of all objects which have any of
    # the given tags.
    query = """
    SELECT %(model_pk)s
    FROM %(model)s, %(tagged_item)s
    WHERE %(tagged_item)s.content_type_id = %(content_type_id)s
      AND %(tagged_item)s.tag_id IN (%(tag_id_placeholders)s)
      AND %(model_pk)s = %(tagged_item)s.object_id
    GROUP BY %(model_pk)s""" % {
        'model_pk': '%s.%s' % (model_table, qn(model._meta.pk.column)),
        'model': model_table,
        'tagged_item': qn(self.model._meta.db_table),
        'content_type_id': ContentType.objects.get_for_model(model).pk,
        'tag_id_placeholders': ','.join(['%s'] * tag_count),
    }

    cursor = connection.cursor()
    cursor.execute(query, [tag.pk for tag in tags])
    object_ids = [row[0] for row in cursor.fetchall()]
    if len(object_ids) > 0:
        return queryset.filter(pk__in=object_ids)
    else:
        return model._default_manager.none()

For part #3 I believe you're looking for an intersection. See line 307 of models.py

def get_intersection_by_model(self, queryset_or_model, tags):
    """
    Create a ``QuerySet`` containing instances of the specified
    model associated with *all* of the given list of tags.
    """
    tags = get_tag_list(tags)
    tag_count = len(tags)
    queryset, model = get_queryset_and_model(queryset_or_model)

    if not tag_count:
        return model._default_manager.none()

    model_table = qn(model._meta.db_table)
    # This query selects the ids of all objects which have all the
    # given tags.
    query = """
    SELECT %(model_pk)s
    FROM %(model)s, %(tagged_item)s
    WHERE %(tagged_item)s.content_type_id = %(content_type_id)s
      AND %(tagged_item)s.tag_id IN (%(tag_id_placeholders)s)
      AND %(model_pk)s = %(tagged_item)s.object_id
    GROUP BY %(model_pk)s
    HAVING COUNT(%(model_pk)s) = %(tag_count)s""" % {
        'model_pk': '%s.%s' % (model_table, qn(model._meta.pk.column)),
        'model': model_table,
        'tagged_item': qn(self.model._meta.db_table),
        'content_type_id': ContentType.objects.get_for_model(model).pk,
        'tag_id_placeholders': ','.join(['%s'] * tag_count),
        'tag_count': tag_count,
    }

    cursor = connection.cursor()
    cursor.execute(query, [tag.pk for tag in tags])
    object_ids = [row[0] for row in cursor.fetchall()]
    if len(object_ids) > 0:
        return queryset.filter(pk__in=object_ids)
    else:
        return model._default_manager.none()