在Django中有没有一种方法可以序列化分页对象？

Question

我正在使用Django分页和jQuery。我可以序列化分页对象的对象列表，但我想序列化整个对象以获取更多数据（页码、总页数...）。 如何序列化整个分页对象？

谢谢

javascript

function getRestaurants(query) {

        $.post("/getRestaurant/", query,
            function(data) {
                /* do stuff with data */
             },"json" );
}

视图.py

def getRestaurant(request):

    results = Restaurant.objects.all()
    paginator = Paginator(restaurants, 5)

    # Make sure page request is an int. If not, deliver first page.
    try:
        page = int(request.POST.get('page','1'))
    except ValueError:
        page = 1

    # If page request (9999) is out of range, deliver last page of results.
    try:
        results = paginator.page(page)
    except (EmptyPage, InvalidPage):
        results = paginator.page(paginator.num_pages)

    data=serializers.serialize("json", results.object_list) #I'd like to serialize the whole results object

    return HttpResponse(data)

- jul

---

我不确定是否有效，但你可以尝试使用http://code.google.com/p/jsonpickle/。 - diegueus9

Answer 1

我刚刚不得不为此提供一个hacky解决方案，所以我将其发布为答案-欢迎任何改进：

from django.core import serializers
from django.core.paginator import Paginator, InvalidPage, EmptyPage
from django.core.serializers.json import DjangoJSONEncoder
from django.http import HttpResponse
from django.utils import simplejson  
from types import MethodType 

from mysite.tasks.models import Task

PER_PAGE = 20

def list(request):
    """
    Return a paginated JSON object.
    """

    paginator = Paginator(tasks.objects.all(), PER_PAGE) 
    try:
        page = int(request.GET.get('page', '1'))
    except ValueError:
        page = 1

    # If page request (9999) is out of range, deliver last page of results.
    try:
        pagetasks = paginator.page(page)
    except (EmptyPage, InvalidPage):
        pagetasks = paginator.page(paginator.num_pages)

    # Dump the Page attributes we want to a dictionary
    serializedpage = {}
    wanted = ("end_index", "has_next", "has_other_pages", "has_previous",
            "next_page_number", "number", "start_index", "previous_page_number")
    for attr in wanted:
        v = getattr(tasks, attr)
        if isinstance(v, MethodType):
            serializedpage[attr] = v()
        elif isinstance(v, (str, int)):
            serializedpage[attr] = v

    # Serialise the queryset to plain Python objects 
    # and add them to the serialized page dictionary
    pythonserializer = serializers.get_serializer("python")()
    serializedpage["object_list"] = pythonserializer.serialize(pagetasks.object_list, 
            fields=('task_id', 'task_data')) 

    # Dump it as JSON using the Django encoder
    response = HttpResponse(mimetype="application/json")
    simplejson.dump(serializedpage, response, cls=DjangoJSONEncoder)
    return response                               

- Mikesname

Answer 2

我的建议是放弃分页器对象，自己切片查询集。这样你就可以轻松地序列化输出。

例如，如果您想将评论列表序列化为JSON格式，可以按如下方式执行：

comment_list = ArticleComment.objects.filter(article__id=int(_id), is_public=True).values('created', 'tag', 'content', 'author').order_by('-created')   
    start = (page_num - 1) * COMMENTS_PER_PAGE
    end = page_num * COMMENTS_PER_PAGE
    return HttpResponse(json.dumps(list(comment_list[start:end]), cls=DjangoJSONEncoder))

- mmbrian

Answer 3

Django Restframework提供了特殊的PaginationSerializer，它可以满足您的需求。

- codingjoe

---

PaginationSerializer在DRF 3.1版本中已被移除。http://stackoverflow.com/questions/31500192/django-rest-framework-error-cannot-import-name-paginationserializer - Michael Scheper