您可以将Q对象用于#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')
)
我相信,工会和交叉点有点超出了Django ORM的范围,但它可能会对这些有所帮助.以下示例来自名为django-tagging的Django应用程序,该应用程序提供了该功能.models.py的第346行:
对于第二部分,基本上你正在寻找两个查询的联合
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()
对于第3部分,我相信你正在寻找一个十字路口.见models.py的第307行
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()
Ycros.. 16
我用Django 1.0测试了这些:
"或"查询:
Blog.objects.filter(tags__name__in=['tag1', 'tag2']).distinct()
或者您可以使用Q类:
Blog.objects.filter(Q(tags__name='tag1') | Q(tags__name='tag2')).distinct()
"和"查询:
Blog.objects.filter(tags__name='tag1').filter(tags__name='tag2')
我不确定第三个,你可能需要放到SQL来做它.
您可以将Q对象用于#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')
)
我相信,工会和交叉点有点超出了Django ORM的范围,但它可能会对这些有所帮助.以下示例来自名为django-tagging的Django应用程序,该应用程序提供了该功能.models.py的第346行:
对于第二部分,基本上你正在寻找两个查询的联合
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()
对于第3部分,我相信你正在寻找一个十字路口.见models.py的第307行
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()
我用Django 1.0测试了这些:
"或"查询:
Blog.objects.filter(tags__name__in=['tag1', 'tag2']).distinct()
或者您可以使用Q类:
Blog.objects.filter(Q(tags__name='tag1') | Q(tags__name='tag2')).distinct()
"和"查询:
Blog.objects.filter(tags__name='tag1').filter(tags__name='tag2')
我不确定第三个,你可能需要放到SQL来做它.
请不要重新发明轮子并使用专为您的用例制作的django-tagging应用程序.它可以完成您描述的所有查询,等等.
如果您需要向Tag模型添加自定义字段,您还可以查看我的django-tagging分支.
这将为你做到这一点
Blog.objects.filter(tags__name__in=['tag1', 'tag2']).annotate(tag_matches=models.Count(tags)).filter(tag_matches=2)
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