如何過濾Django QuerySet的相關字段'all'或'none'

Question

E.g我有一個模型Cat，它有一個與Life相反的ForeignKey。

class Life(models.Model): 
    state = models.CharField(choices=('alive', 'dead', 'unknown') 
    cat = models.ForeignKey('animals.Cat', related_name="lives") 


class Cat(models.Model): 
    name = models.CharField(max_length=12) 
    cat_type = models.CharField(choices=('normal', 'schroedinger') 
    ... 

如何獲得Cat的QuerySet A S已沒有失去他們的生活？即有所有他們的生活無論是在國家「活着」或者是cat_type「薛定諤」，並有沒有自己的生活狀態「死」）

Answer 1

我的天堂「T使用一段時間這個API，但我相信這會完成這項工作：

from django.db.models import Q 

normal_and_alive = Q(cat_type="normal") & ~Q(lives__state__in=["dead", "unknown"]) 
schroedinger_and_not_dead = Q(cat_type="schroedinger") & ~Q(lives__state="dead") 

cats = Cat.objects.filter(normal_and_alive | schroedinger_and_not_dead) 

見Django的文檔的文檔爲complex lookups with the Q() object

撇開：這將只執行一個數據庫查詢

Answer 2

cats = Cat.objects.filter(id=-1) # make it an empty queryset 

temp = Cat.objects.filter(cat_type='normal') 
for one in temp: 
    cats |= one.lives.filter(state='alive') # set union 

temp = Cat.objects.filter(cat_type='schroedinger') 
for one in temp: 
    cats |= one.lives.exclude(state='dead') 

return cats 

新的答案：

cats = Cat.objects.filter(id=-1) # make it an empty queryset 

temp = Cat.objects.filter(cat_type='normal') 
for one in temp: 
    if len(one.lives.exclude(state='alive')) == 0: 
     cats |= Cat.objects.filter(id=one.id) 

temp = Cat.objects.filter(cat_type='schroedinger') 
for one in temp: 
    if len(one.lives.filter(state='dead')) == 0: 
     cats |= Cat.objects.filter(id=one.id) 

return cats