Question

我正在开发一个django应用程序，我可以根据食谱过滤成分。我正在使用django过滤器来为我的用户提供过滤选项。我的过滤下拉工作完全正常，但我想在提交时添加一个选项，所有成分应该列出，无论他们的食谱如何。

这是我的代码：

#models.py

class Recipe(models.Model):
    user = models.ForeignKey('auth.User')
    title = models.CharField(max_length=500)
    description = models.TextField(max_length=500)
    rules = models.TextField(max_length=500,blank=True)
    def __str__(self):
            return self.title
class Ingredient(models.Model):
    user = models.ForeignKey('auth.User')
    recipe_id = models.ForeignKey(Recipe, on_delete=models.CASCADE)
    title = models.CharField(max_length=500)
    instructions = models.CharField(max_length=500)
    rules = models.TextField(max_length=500,blank=True)
    primal = models.CharField(default='0',max_length=500,blank=True)
    def __str__(self):
            return self.title

#forms.py

class RecipeFilter(django_filters.FilterSet):
    class Meta:
        model = Ingredient
        fields = ['recipe_id']

#views.py

def ingredient_list(request):
    ingredientfilter = IngredientFilter( queryset=Recipe.objects.filter(user = request.user))
    if request.method == 'GET' and 'recipe_id' in request.GET:
        recipe_id=request.GET['recipe_id'];
        ingredients = Ingredient.objects.filter(recipe_id= recipe_id)
        selected_combo_value = Recipe.objects.get(pk=recipe_id)
        return render(request, 'ingredient_list.html',{'ingredients':ingredients, 'ingredientfilter': ingredientfilter,'selected_combo_value':selected_combo_value })
    else:
        ingredients = Ingredient.objects.filter(user = request.user)
    return render(request, 'ingredient_list.html',{'ingredients':ingredients, 'ingredientfilter': ingredientfilter })

知道该怎么做吗？

Answer 1

这是一个众所周知的issue。解决方法是覆盖过滤器的选项。

class RecipeFilter(django_filters.FilterSet):
    def __init__(self, *args, **kwargs):
        super(RecipeFilter, self).__init__(*args, **kwargs)
        self.filters['recipe_id'].field.choices.insert(0, ('', u'---------'))

如何制作一个选项＆＃39; All＆＃39;在过滤数据时在django过滤器中

1 个答案: