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在 Django 1.4.2 中，我使用了官方文档中给出的简单分页代码：

...
paginator = Paginator(songs, 25) # Show 25 songs per page

page = request.GET.get('page')
try:
    songs = paginator.page(page)
except PageNotAnInteger:
    # If page is not an integer, deliver first page.
    songs = paginator.page(1)
except EmptyPage:
    # If page is out of range (e.g. 9999), deliver last page of results.
    songs = paginator.page(paginator.num_pages)
...

当我运行该视图时，我在线收到“ 'Manager' object is unsubscriptable ”错误songs = paginator.page(1)。我搜索了错误，但找不到任何解决方案。问题是什么？

编辑：

songs是 django 模型对象列表。全视图代码是这样的：

def index(request):
    songs = Song.objects

    #filter params
    q_name = request.GET.get('name', None)
    if q_name:
        songs = songs.filter(name__contains=q_name)

    q_composer = request.GET.get('composer', None)
    if q_composer:
        songs = songs.filter(composer__name__contains=q_composer)

    q_composer_id = request.GET.get('composer_id', '')
    if q_composer_id != '':
        songs = songs.filter(composer__id=q_composer_id)

    paginator = Paginator(songs, 25) # Show 25 contacts per page

    page = request.GET.get('page')
    try:
        songs = paginator.page(page)
    except PageNotAnInteger:
        # If page is not an integer, deliver first page.
        songs = paginator.page(1)
    except EmptyPage:
        # If page is out of range (e.g. 9999), deliver last page of results.
        songs = paginator.page(paginator.num_pages)

    return render(request, 'index.html', {'songs': songs})
4

1 回答 1

1

当没有提供过滤器时会发生什么。然后，您尝试对 AuditManger 而不是查询集进行分页。您应该将第一个查询更改为：

songs = Song.objects.all()
于 2012-12-05T20:23:32.267 回答