Django ORM展示多关联对象问题：仅显示部分Contract对象求助

问题背景 #

我尝试用Django ORM展示多个关联对象，原代码逻辑如下：

if co is None:
    try:
      address = Contract.objects.filter(address= servicesearch)
      print(address)
    except Contract.DoesNotExist:
      address= None

    for add in address:
      co = add.contractservices_set.all()
      print(co)

首次查询得到QuerySet：

return (<QuerySet [<Contract: Contract object (52)>, <Contract: Contract object (55)>]>
)

循环查询得到两个Contractservices的QuerySet：

return (<QuerySet [<Contractservices: Contractservices object (88)>, <Contractservices: Contractservices object (89)>, <Contractservices: Contractservices object (90)>, <Contractservices: Contractservices object (91)>]>
<QuerySet [<Contractservices: Contractservices object (98)>, <Contractservices: Contractservices object (101)>, <Contractservices: Contractservices object (102)>, <Contractservices: Contractservices object (99)>, <Contractservices: Contractservices object (103)>, <Contractservices: Contractservices object (100)>, <Contractservices: Contractservices object (104)>]>
)

查询结果正常，但渲染时仅显示Contract对象55的内容，我需要同时展示52和55这两个Contract对象。分页器代码如下：

paginator = Paginator(co,8)
if request.GET.get('page') == None:
    page_number = 1
else:
    page_number =request.GET.get('page')
  
DataFinal=paginator.get_page(page_number)
totalpage =DataFinal.paginator.num_pages

且无法修改模板，因为这只是众多搜索条件之一，需要解决该问题。

问题原因 #

循环中每次给co赋值时，都会覆盖之前的QuerySet，最终co仅保留了最后一个Contract（id=55）的contractservices数据，导致分页和渲染时只显示这部分内容。

解决方案 #

方法1：合并所有关联对象QuerySet #

通过收集所有Contract对应的contractservices，再合并为一个可迭代对象（兼容分页器）：

from itertools import chain

if co is None:
    try:
        address = Contract.objects.filter(address=servicesearch)
        print(address)
    except Contract.DoesNotExist:
        address = None

    # 收集所有Contract对应的contractservices
    services_list = []
    for add in address:
        services_list.append(add.contractservices_set.all())
    # 合并为一个列表（分页器支持列表类型）
    co = list(chain(*services_list))

方法2：反向查询（更高效，避免N+1问题） #

直接从Contractservices模型反向筛选关联Contract符合条件的记录，仅需1次数据库查询：

if co is None:
    # 直接查询所有属于address匹配的Contract的Contractservices
    co = Contractservices.objects.filter(contract__address=servicesearch)
    print(co)

两种方法修改后，co都会包含Contract 52和55对应的所有Contractservices数据，分页渲染时即可展示全部内容，且无需修改现有模板和分页代码。

内容的提问来源于stack exchange，提问作者Arghya Dasmahapatra