對象過濾器 - 需要提醒

Question

比方說，我有三個型號：

class ONE (models.Model): 
    [some awesome stuff] 

class TWO (models.Model): 
    one = models.ForeignKey(ONE) 
    user = models.ForeignKey(User) 
    [other stuff] 

class THREE (models.Model): 
    one = models.ForeignKey(ONE) 
    [stuff and whatnot] 

的看法目前看起來是這樣的：

def my_view(request, username, template_name='view.html'): 
    current_user = get_object_or_404(User, username=username) 
    two = TWO.objects.filter(user_id=current_user) 
    three = THREE.objects.filter(one_id=two) 

    data = { 
     'user': current_user, 
     'two': two, 
     'three': three 
    } 

    return render(request, template_name, data) 

數據庫條目（它被簡化;有更多的數據在每個表中）：

Table ONE 
---------- 
id=1 
id=2 
id=3 

Table TWO 
---------- 
one_id=1; user_id=1 
one_id=2; user_id=1 
one_id=3; user_id=1 

Table THREE 
---------- 
one_id=1 
one_id=2 
one_id=3 

有了上述觀點，我不能夠顯示錶THREE中來自表TWO的user_id = 1的所有條目。

錯誤消息= more than one row returned by a subquery used as an expression

我試圖用three = THREE.objects.filter(one_id__in=two)但不顯示HTML頁面上的任何內容。沒有錯誤信息。

HTML：

{% for two in two %} 
    <p>{{ two.one }}</p> <!-- that works! --> 
{% endfor %} 
{% for three in three %} 
<ul> 
    <li>{{ three }}</li> 
</ul> 
{% endfor %} 

是否有我需要的過濾器選項？

Answer 1

嘗試更正__in過濾器。該參數必須是list實例，而不是QuerySet：

two = TWO.objects.filter(user_id=current_user) 
two_ids = two.values_list('id', flat=True).distinct('id') 
three = THREE.objects.filter(one_id__in=two_ids) 

distinct這裏不是必需的，但強烈推薦。

多見於https://docs.djangoproject.com/en/dev/ref/models/querysets/#django.db.models.query.QuerySet.values_list https://docs.djangoproject.com/en/dev/ref/models/querysets/#django.db.models.query.QuerySet.distinct