Question

这是我的代码。

obj_list=Location.objects.all()
first_element=obj_list[0]
last_element=obj_list[-1]

然后，

return render_to_response(template_name, {
        'first_element':first_element,
        'last_element':last_element,
    })

并在模板中：

{{ first_element.terminal_id}} {{last_element.terminal_id}}

但它没有显示任何内容，

我能做什么，

感谢

Answer 1

查看http://docs.djangoproject.com/en/dev/topics/db/queries/#limiting-querysets

不支持负索引（即Entry.objects.all()[-1]）。

尝试：

first_element = Location.objects.all()[0]
last_element = Location.objects.all().reverse()[0]

- 更新8/6/17 -

基于@MisterRios的评论，

从1.6开始，Django支持在查询集上使用.first()和.last()： first_element = Location.objects.first() last_element = Location.objects.last()

参考：https://docs.djangoproject.com/en/1.7/ref/models/querysets/#django.db.models.query.QuerySet.first

Answer 2

您可能无法对查询集进行否定索引，但您可以将该查询集放入列表中，然后将其索引。

locations = list(Location.objects.all())
first_element = locations[0]
last_element = locations[-1]

但这种效率非常低，并且只应在表中存在少量位置时使用，并且您希望保持代码简单。否则，如果确实需要提高效率，请参阅@ pterk的答案，涉及聚合和Min / Max。

Answer 3

要获取最后一个[-1]，请尝试Location.objects.latest('id')，如文档中所示：

https://docs.djangoproject.com/en/1.3/ref/models/querysets/#latest

Answer 4

最后： - Location.objects.reverse()[0]

或

Location.objects.all()[Location.objects.count()-1] // BAD WAY

首先： Location.objects.all()[0]

注意：不支持负索引。所以，Location.objects.all()[-1]会给你一个 AssertionError

Answer 5

如果有人来这里获取相关模型的第一个和最后一个项目 - 有效地做到这一点的唯一方法是将相关字段转换为列表或使用count（）来获取最后一个项目的索引（使用Django 1.11.2 ）：

class Parent(models.Model):
    name = models.CharField(max_length=200)


class Child(models.Model):
    parent = models.ForeignKey(Parent, on_delete=models.CASCADE, related_name='children')
    name = models.CharField(max_length=200)


class ParentTest(TestCase):
    def test_parent(self):
        # create some data
        for p in range(10):
            parent = Parent.objects.create(name=p)
            for c in range(10):
                parent.children.create(name=c)

        with self.assertRaises(AssertionError):  # can't negative index
            parents = Parent.objects.prefetch_related('children')
            for parent in parents:
                first = parent.children.all()[0]
                last = parent.children.all()[-1]

        with self.assertNumQueries(22):  # 2 for prefetch and 20 for access
            parents = Parent.objects.prefetch_related('children')
            for parent in parents:
                first = parent.children.first()
                last = parent.children.last()

        with self.assertNumQueries(22):  # 2 for prefetch and 20 for access
            parents = list(Parent.objects.prefetch_related('children'))
            for parent in parents:
                first = parent.children.first()
                last = parent.children.last()

        with self.assertNumQueries(12):  # 2 for prefetch and 10 for last
            parents = Parent.objects.prefetch_related('children')
            for parent in parents:
                first = parent.children.all()[0]
                last = parent.children.reverse()[0]

        with self.assertRaises(AssertionError):  # can't negative index
            parents = list(Parent.objects.prefetch_related('children'))
            for parent in parents:
                first = parent.children.all()[0]
                last = parent.children.all()[-1]

        with self.assertNumQueries(2):  # 2 for prefetch
            parents = Parent.objects.prefetch_related('children')
            for parent in parents:
                children = list(parent.children.all())
                first = children[0]
                last = children[-1]

        with self.assertNumQueries(2):
            parents = Parent.objects.prefetch_related('children')
            for parent in parents:
                first = parent.children.all()[0]
                last = parent.children.all()[parent.children.count() - 1]

Answer 6

效率可能很低

obj_list=Location.objects.all()
first_element=obj_list[0]
last_element=obj_list[len(obj_list)-1]

Answer 7

如果您有办法对位置对象进行排序，请查看聚合（最小和最大）。 http://docs.djangoproject.com/en/dev/topics/db/aggregation/

您可能会在id上尝试Min和Max，但请尽量避免这种情况，因为ID的顺序不是保证（至少不会跨越各种数据库引擎）

Answer 8

obj_list 是 QuerySet 的一个实例，QuerySet 有 own methods：

obj_list.latest('pk')
obj_list.earliest('pk')
obj_list.first() 
obj_list.last()

Answer 9

问题查询：

and


ERROR 2021-05-17 02:14:20,744 log 30 139680490260288 Internal Server Error: /camp/
Traceback (most recent call last):
  File "/usr/local/lib/python3.7/site-packages/django/core/handlers/exception.py", line 34, in inner
    response = get_response(request)
  File "/usr/local/lib/python3.7/site-packages/django/core/handlers/base.py", line 115, in _get_response
    response = self.process_exception_by_middleware(e, request)
  File "/usr/local/lib/python3.7/site-packages/django/core/handlers/base.py", line 113, in _get_response
    response = wrapped_callback(request, *callback_args, **callback_kwargs)
  File "/usr/local/lib/python3.7/site-packages/django/contrib/auth/decorators.py", line 21, in _wrapped_view
    return view_func(request, *args, **kwargs)
  File "/usr/local/lib/python3.7/site-packages/django/contrib/auth/decorators.py", line 21, in _wrapped_view
    return view_func(request, *args, **kwargs)
  File "/opt/project/apps/web/views.py", line 419, in camp
    amount_by_daily = draw_amount_by_daily()
  File "/opt/project/apps/web/charts.py", line 16, in draw_amount_by_daily
    daily_list = DailyReport.objects.filter(category="daily_all").order_by('create_at').values("create_at")[-12:]
  File "/usr/local/lib/python3.7/site-packages/django/db/models/query.py", line 288, in __getitem__
    "Negative indexing is not supported."
AssertionError: Negative indexing is not supported.

如何解决？

daily_list = DailyReport.objects.filter(category="daily_all").order_by('create_at').values("create_at")[-12:]

提示：

daily_list = DailyReport.objects.filter(category="daily_all").order_by('-create_at').values("create_at")[:12]
result = list(reversed(daily_list))

如何使用django获取第一个元素和最后一个元素，Location.objects.all（）

9 个答案:

问题查询：

如何解决？

提示：