django - 'NoneType' 类型的参数对于 django 字符串“”是不可迭代的

Question

我正在尝试根据空格拆分我的查询" "。示例first_name last_name但是，如果我尝试检查搜索中的空格，则会收到错误消息

“NoneType”类型的参数不可迭代

if ' ' in query: #This is where is the error is generated

试图解决上述问题。下面是整个代码

class QList(SelectRelatedMixin, ListView):
    model = Question

    def get_queryset(self):
        queryset = super(QList, self).get_queryset().order_by('-created_at')
        query = self.request.GET.get('q')

        if ' ' in query:
            query = query.split()
            queryset = queryset.filter(
                chain(User.objects.filter(first_name__icontains=query[0], last_name__icontains=query[1]),
                      User.objects.filter(first_name__icontains=query[1], last_name__icontains=query[0])))

            return queryset
        else:
            queryset = queryset.filter(
                Q(user__username__iexact=query) |
                Q(user__first_name__iexact=query) |
                Q(user__last_name__iexact=query)

                )
        return queryset

任何人都可以建议一个解决方法

标签： djangopython-3.xdjango-viewsdjango-queryset

Answer 1

要拆分字符串，只需说你的名字是“Marco Bianchi”。

query = "Marco Bianchi"
first_name = query.split(" ")[0]
last_name = query.split(" ")[1]

更新：而不是

if ' ' in query:
     query = query.split()

用这个

query = query.split(" ")