在 Django 中获取下一个和上一个对象

Question

我正在尝试获取漫画书问题的下一个和上一个对象。简单地更改 ID 号或通过添加日期进行过滤是行不通的，因为我没有按顺序添加问题。

这就是我的视图的设置方式，它适用于 prev_issue 并返回前一个对象，但它返回 next_issue 的最后一个对象，我不知道为什么。

def issue(request, issue_id):
    issue = get_object_or_404(Issue, pk=issue_id)
    title = Title.objects.filter(issue=issue)
    prev_issue = Issue.objects.filter(title=title).filter(number__lt=issue.number)[0:1]
    next_issue = Issue.objects.filter(title=title).filter(number__gt=issue.number)[0:1]

Answer 1

注意不要使用object.get(pk=object.pk + 1)这些东西，IntegrityError如果该pk处的对象被删除，则始终使用查询集

对于访客：

''' Useage '''
"""
    # Declare our item
    store = Store.objects.get(pk=pk)
    # Define our models
    stores = Store.objects.all()
    # Ask for the next item
    new_store = get_next_or_prev(stores, store, 'next')
    # If there is a next item
    if new_store:
        # Replace our item with the next one
        store = new_store
"""

''' Function '''
def get_next_or_prev(models, item, direction):
    '''
    Returns the next or previous item of
    a query-set for 'item'.

    'models' is a query-set containing all
    items of which 'item' is a part of.

    direction is 'next' or 'prev'
    
    '''

    getit = False
    if direction == 'prev':
        models = models.reverse()
    for m in models:
        if getit:
            return m
        if item == m:
            getit = True
    if getit:
        # This would happen when the last
        # item made getit True
        return models[0]
    return False

original author

用法

# you MUST call order by to pass in an order, otherwise QuerySet.reverse will not work
qs = Model.objects.all().order_by('pk')
q = qs[0]
prev = get_next_or_prev(qs, q, 'prev')
next = get_next_or_prev(qs, q, 'next')

Answer 2

next_obj_id = int(current_obj_id) + 1 next_obj = Model.objects.filter(id=next_obj_id).first()

prev_obj_id= int(current_obj_id) - 1 prev_obj = Model.objects.filter(id=prev_obj_id).first()

#你没有什么可以放松的......这对我有用

Answer 3

from functools import partial, reduce
from django.db import models


def next_or_prev_instance(instance, qs=None, prev=False, loop=False):

    if not qs:
        qs = instance.__class__.objects.all()

    if prev:
        qs = qs.reverse()
        lookup = 'lt'
    else:
        lookup = 'gt'

    q_list = []
    prev_fields = []

    if qs.query.extra_order_by:
        ordering = qs.query.extra_order_by
    elif qs.query.order_by:
        ordering = qs.query.order_by
    elif qs.query.get_meta().ordering:
        ordering = qs.query.get_meta().ordering
    else:
        ordering = []

    ordering = list(ordering)

    if 'pk' not in ordering and '-pk' not in ordering:
        ordering.append('pk')
        qs = qs.order_by(*ordering)

    for field in ordering:
        if field[0] == '-':
            this_lookup = (lookup == 'gt' and 'lt' or 'gt')
            field = field[1:]
        else:
            this_lookup = lookup
        q_kwargs = dict([(f, get_model_attr(instance, f))
                         for f in prev_fields])
        key = "%s__%s" % (field, this_lookup)
        q_kwargs[key] = get_model_attr(instance, field)
        q_list.append(models.Q(**q_kwargs))
        prev_fields.append(field)
    try:
        return qs.filter(reduce(models.Q.__or__, q_list))[0]
    except IndexError:
        length = qs.count()
        if loop and length > 1:
            return qs[0]
    return None


next_instance = partial(next_or_prev_instance, prev=False)
prev_instance = partial(next_or_prev_instance, prev=True)

Answer 4

添加一个order_by 子句以确保它按number 排序。

next_issue = Issue.objects.filter(title=title, number__gt=issue.number).order_by('number').first()

Answer 5

如果需要查找按可以相等的字段值排序的下一个和上一个对象，并且这些字段不属于Date*类型，查询变得稍微复杂，因为：

• 对受[:1] 限制的具有相同值的对象进行排序将始终为多个对象产生相同的结果；
• 对象本身可以包含在结果集中。

这里的查询集还考虑了产生正确结果的主键（假设来自 OP 的 number 参数不是唯一的，并且省略了 title 参数，因为它与示例无关）：

上一个：

prev_issue = (Issue.objects
    .filter(number__lte=issue.number, id__lt=instance.id)
    .exclude(id=issue.id)
    .order_by('-number', '-id')
    .first())

下一步：

next_issue = (Issue.objects
    .filter(number__gte=issue.number, id__gt=instance.id)
    .exclude(id=issue.id)
    .order_by('number', 'id')
    .first())

Answer 6

我知道这有点晚了，但对于其他人来说，django 确实有更好的方法来做到这一点，请参阅https://docs.djangoproject.com/en/1.7/ref/models/instances/#django.db.models.Model.get_previous_by_FOO

所以这里的答案应该是

next_issue = Issue.get_next_by_number(issue, title=title)

Django 管理器用一点元类的技巧来做到这一点。