Django 如何注释嵌套 forloop 查询集字典的计数

Django how to annotate count nested forloop queryset dictionary

美好的一天,

我的问题是:

我的项目取决于对做出任何操作的用户给予积分,例如(post、评论、收藏、喜欢......)。

因此,在用户列表页面中,我想列出所有用户和每个用户的其他数据(姓名、积分、徽章......)

为了给用户积分,我必须计算他的post、评论、喜欢等等.....

我尝试了几种方法和方式,但都无法获得注释或prefetch_related或select_related

Models.py

class Post(models.Model):
    title = models.CharField(max_length=100)
    content = models.TextField()
    author = models.ForeignKey(User, on_delete=models.CASCADE, related_name='posts')

Views.py

def user_list(request):
    users = User.objects.all()
    template = 'user/users_list.html'

    nested_posts = {}
    for user in users:
        posts = user.posts.all()
        nested_posts[user, posts] = posts.count()

    print("nested : ", nested_posts)

    context = {
        'users': users,
        'user':user,
        'posts': posts,
        'nested_posts': nested_posts,}
    return render(request, template, context)

当我打印 nested .. 我发现每个用户的计数 posts .. 但是我怎样才能在计算字段 re-use 中将其作为变量

查询集

nested :  {(<User: Fareed>, <QuerySet [<Post: Senior Purchasing Specialist>]>): 1, 
(<User: Hussein>, <QuerySet [<Post: Senior Software Development Engineer in Test>]>): 1, 
(<User: Karima>, <QuerySet []>): 0, 
(<User: Yahia>, <QuerySet []>): 0}

我也试过了:

GetUserID = User.objects.get(id=2)
var01 = GetUserID.posts.all().count()

但这是针对一个用户 (id=2) .. 并且确保所有用户都获得了用户 (id=2) 的总 posts,而不是每个用户。

我也试过了:

Posts_count_per_user = User.posts.annotate(posts_count=Count('posts'))
User_Score_of_posts = Posts_count_per_user.aggregate(posts_score=Count('posts_count') * 1000)

但是我得到了这个错误:

'ReverseManyToOneDescriptor' object has no attribute 'annotate'

有什么建议请...

提前致谢,

你试过这个吗,

user_qs = User.objects.annotate(posts_count=Count('posts'))

# usage
for user_instance in user_qs:
    print("post count: ", user_instance.posts_count)
    print("post score: ", user_instance.posts_count * 1000)

或者您可以在数据库级别本身注释 post 分数,

from django.db.models import F, Count

user_qs = User.objects.annotate(<b>posts_count=Count('posts'), posts_score=F('posts_count') * 1000</b>)