过滤 Django 查询集

Question

在一个项目中，我有一个视图，它显示在浏览器中作为搜索栏来搜索食谱。我认为有比我现在做的更好的方法，但不知道如何做。如果我现在不做我正在做的事情，我会收到您无法使用 None 过滤的错误。

我知道我可以使用 try except 但没有任何值可能是 None 或引发错误。

class SearchResultListViewOther(ListView):
    model = Recipe
    template_name = 'recipes/search_other.html'
    extra_context = {
        'time': time,
        'categorie': categorie,
        'continent': continent,
    }

    def get_queryset(self):
        """
        Filter out recipes by some other params like the title
        """
        # Can not filter on None so set a value if no search params were given
        title = self.request.GET.get('title')
        if not title:
            title = '0'
        time_recipe = self.request.GET.get('time')
        if not time_recipe:
            time_recipe = 0
        continent_recipe = self.request.GET.get('continent')
        if not continent_recipe:
            continent_recipe = '0'
        object_list = Recipe.objects.filter(
            Q(title__icontains=title) | Q(time__lte=time_recipe) |
            Q(continent__exact=continent_recipe)
            )
        return object_list

Answer 1

您可以使用&= 和|= 操作链接Q 查询。

def get_queryset(self):
    """
    Filter out recipes by some other params like the title
    """
    title = self.request.GET.get('title')
    time_recipe = self.request.GET.get('time')
    continent_recipe = self.request.GET.get('continent')

    # default empty query
    q = Q()
    if title:
        q |= Q(title__icontains=title)
    if time_recipe:
        q |= Q(time__lte=time_recipe)
    if continent_recipe:
        q |= Q(continent__exact=continent_recipe)
    return  Recipe.objects.filter(q)

Answer 2

def get_queryset(self):
    """
    Filter out recipes by some other params like the title
    """
    # Can not filter on None so set a value if no search params were given
    title = self.request.GET.get('title')
    time_recipe = self.request.GET.get('time')
    continent_recipe = self.request.GET.get('continent')
    if title or time_recipe or continent_recipe:
        return  Recipe.objects.filter(
        Q(title__icontains=title) | Q(time__lte=time_recipe) |
        Q(continent__exact=continent_recipe)
        )

请试试这个。