Как получить общее количество данных, созданных во многих областях

Question

здесь, в этом я хочу получить ни одного из Players_Participants, добавленных пользователем, в этом я получаю общее количество Players_Participants

class MyUser(AbstractBaseUser):

    user_name=models.CharField(max_length=10,blank=True,null=True,unique=True)

    def __str__(self):
        return str(self.user_name)  

class Session(models.Model):

    Host=models.ForeignKey(MyUser,on_delete=models.CASCADE,related_name='host')
    game=( 
        ('cricket','cricket'),
        ('football','football'),
        ('basketball','basketball'),
        ('hockey','hockey'),
        ('gym','gym'),
        ('baseball','baseball'),
    )

    Sport=models.CharField(max_length=20,choices=gamee)      
    Players_Participating=models.ManyToManyField(MyUser,related_name='related')

    def __str__(self):
        return str(self.Host)

class MyUserViewSet(viewsets.ViewSet):

    def create(self,request):
        user_name=request.data.get('user_name')
        user=MyUser()
        user.user_name=user_name
        return Response({"response":'delivered'})

class SessionViewSet(viewsets.ViewSet):   

    def create(self, request):
        Host= request.data.get('Host')
        Sport = request.data.get('Sport')
        Players_Participating=request.data.get('Players_Participating')
        new=Session()
        new.Host=MyUser.objects.get(user_name=Host)
        new.Sport=Sport
        new.Players_Participating=Players_Participating
        new.save()

    def list(self,request):
        users = Session.objects.all()
        use = []
        player_list=[]
        for user in users:
            for players in user.Players_Participating.all():
                player_list.append(players.user_name)
            use.append({
                    'Host':user.Host.user_name,
                    "Sport":user.Sport,
                    "Players_Participating":player_list,        
                    })
        return Response({"success":True, "users":use, "pages":paginator.num_pages})