Question

我正在使用Django进行简单搜索，并使用分页中的构建对结果进行分页，当我单击“下一步”以查看下一页结果时出现此错误，int（）参数必须是字符串，类似于字节对象或数字，而不是“ NoneType” 我不知道是什么原因造成的，这是代码，q是查询字符串

def result(request):
    try:
        q = request.GET.get('q')
        orders = Order.objects.filter(Q(flat_number=int(q)) | Q(customer_contact=int(q)))
        paginator = Paginator(orders, 10) # Show 25 contacts per page
        page = request.GET.get('page')
        orders = paginator.get_page(page)
        context['orders'] = orders
        return render(request, 'main/results.html',context)
    except Exception as e:
        print('error is ', e)
        return HttpResponse(str(e))

Answer 1

在request.GET.get('q')中，如果q没有作为get调用中的参数传递，它将返回None。

如果未传递q，将其更改为request.GET['q']将引发异常

Answer 2

这里是如何使用分页器模块（用于空白页面等）的示例。

from django.core.paginator import Paginator, EmptyPage, PageNotAnInteger


def post_list(request): #list items
    queryset_list = Post.objects.all() #.order_by("-timestamp")

    query = request.GET.get("q")
    if query:
        queryset_list = queryset_list.filter(
            Q(title__icontains=query) |
            Q(content__icontains=query)


            ).distinct()

    paginator = Paginator(queryset_list, 10) # Show 25 contacts per page
    page_request_var = "page"
    page = request.GET.get(page_request_var)
    try:
        queryset = paginator.page(page)
    except PageNotAnInteger:
        # If page is not an integer, deliver first page.
        queryset = paginator.page(1)
    except EmptyPage:
        # If page is out of range (e.g. 9999), deliver last page of results.
        queryset = paginator.page(paginator.num_pages)
    context = {
        "object_list": queryset,
        "title": "Building stuff with code",
        "page_request_var" : page_request_var
    }
    return render(request, "post_list.html", context)

Answer 3

以以下方式尝试

def result(request):
    try:
        q = request.GET.get('q')
        orders = Order.objects.filter(Q(flat_number=int(q)) | 
         Q(customer_contact=int(q)))
        paginator = Paginator(orders, 10) # Show 25 contacts per page
        page = request.GET.get('page')
        try:
            pots = paginator.page(page)
        except PageNotAnInteger:
            posts = paginador.page(1)
        except EmptyPage:
            posts = paginador.page(1)
        context['orders'] = post
        return render(request, 'main/results.html',context)


    except Exception as e:
        print('error is ', e)
        return HttpResponse(str(e))

Django Paginator页面返回NoneType

3 个答案: