之前同事在项目中写了分页的一个函数,但是并没有返回结果集的总个数和总页数。所以我就想到了用 django 自带的分页类获取分页的数据。因为要分页的对象可能是个列表而不是 django 模型的查询集。只是使用了Paginator类查看总页数和总个数的方法。
from django.core.paginator import Paginator
page_rows= "每页展示多少条数据"
# 注queryset是一个模型的查询集
p= Paginator(queryset, page_rows)
# 获取查询集的总个数
total_count= p.count
# 获取查询集的总页数
total_page= p.num_pages
Paginator类源码
class Paginator:
def __init__(self, object_list, per_page, orphans=0,
allow_empty_first_page=True):
self.object_list = object_list
self._check_object_list_is_ordered()
self.per_page = int(per_page)
self.orphans = int(orphans)
self.allow_empty_first_page = allow_empty_first_page
def validate_number(self, number):
"""Validate the given 1-based page number."""
try:
if isinstance(number, float) and not number.is_integer():
raise ValueError
number = int(number)
except (TypeError, ValueError):
raise PageNotAnInteger(_('That page number is not an integer'))
if number < 1:
raise EmptyPage(_('That page number is less than 1'))
if number > self.num_pages:
if number == 1 and self.allow_empty_first_page:
pass
else:
raise EmptyPage(_('That page contains no results'))
return number
def get_page(self, number):
"""
Return a valid page, even if the page argument isn't a number or isn't
in range.
"""
try:
number = self.validate_number(number)
except PageNotAnInteger:
number = 1
except EmptyPage:
number = self.num_pages
return self.page(number)
def page(self, number):
"""Return a Page object for the given 1-based page number."""
number = self.validate_number(number)
bottom = (number - 1) * self.per_page
top = bottom + self.per_page
if top + self.orphans >= self.count:
top = self.count
return self._get_page(self.object_list[bottom:top], number, self)
def _get_page(self, *args, **kwargs):
"""
Return an instance of a single page.
This hook can be used by subclasses to use an alternative to the
standard :cls:`Page` object.
"""
return Page(*args, **kwargs)
@cached_property
def count(self):
"""Return the total number of objects, across all pages."""
c = getattr(self.object_list, 'count', None)
if callable(c) and not inspect.isbuiltin(c) and method_has_no_args(c):
return c()
return len(self.object_list)
@cached_property
def num_pages(self):
"""Return the total number of pages."""
if self.count == 0 and not self.allow_empty_first_page:
return 0
hits = max(1, self.count - self.orphans)
return ceil(hits / self.per_page)
@property
def page_range(self):
"""
Return a 1-based range of pages for iterating through within
a template for loop.
"""
return range(1, self.num_pages + 1)
def _check_object_list_is_ordered(self):
"""
Warn if self.object_list is unordered (typically a QuerySet).
"""
ordered = getattr(self.object_list, 'ordered', None)
if ordered is not None and not ordered:
obj_list_repr = (
'{} {}'.format(self.object_list.model, self.object_list.__class__.__name__)
if hasattr(self.object_list, 'model')
else '{!r}'.format(self.object_list)
)
warnings.warn(
'Pagination may yield inconsistent results with an unordered '
'object_list: {}.'.format(obj_list_repr),
UnorderedObjectListWarning,
stacklevel=3
)
我自定义我的分页函数只是用了 count 和 num_pages 方法,因为我初始化 Paginator 时传入的 queryset 没有进行排序,就触发了 _check_object_list_is_ordered方法的警告。官方的解释是 【A list, tuple, QuerySet, or other sliceable object with a count() or __len__() method. For consistent pagination, QuerySets should be ordered, e.g. with an order_by() clause or with a default ordering on the model】.在一个 qq 群里有人给我解释分页必须要保证幂等,换句话说是我第一页的内容不管查多少次,都是第一页的内容。在这种情况下幂等是交由 order_by 的数据来保证的,在常见数据库中,保持一个序列顺序固定,是需要显示的 order by 来做,虽然我们查询的时候,默认会根据自增 ID 来做一次 order by,但是这是一个不可靠行为,或者说是叫做 undefined behavior ,Django 为了保证在不同数据库,不同版本的数据库中数据一致,加上了这样一个强制。
我还想说一句,num_pages 方法中的ceil函数使用让我眼前一亮,以前求总页数我都是用数学运算分好几种情况考虑,但是看了源码,让人眼前一亮,真的很厉害!
def ceil(*args, **kwargs): # real signature unknown
"""
Return the ceiling of x as an Integral.
This is the smallest integer >= x.
"""
pass
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