我想註解一個對象列表(每個對象都有一組tags
和一個整數值points
),以便它們可以按照標記加上點的總數進行排序。如何通過總共兩個字段對django queryset進行排序?
但是,我找不到任何方式來註釋與兩列的總和對象。任何想法,我應該如何做到這一點?
這些是我正在使用的模型,leaderboard()
函數是我難以工作的函數。
class Game (models.Model):
users = models.ManyToManyField(User, through='Player', related_name='games')
def leaderboard (self):
""" Return a list of players ranked by score, where score is the total of the count of the players tags and their points. """
leaderboard = self.player_set.annotate(
tagcount=models.Count('tags')
).extra(
select={'score': 'tagcount + points'},
order_by=('score',)
)
return leaderboard
class Player (models.Model):
game = models.ForeignKey(Game)
user = models.ForeignKey(User)
points = models.SmallIntegerField(default=0, help_text="Points that have been awarded to the player")
class Tag (models.Model):
game = models.ForeignKey(Game, related_name='tags')
player = ForeignKey(Player, related_name='tags')
編輯2:使用額外的
好了,所以我得到了額外的工作,通過手動計數的標籤數量,並稱該點解決方案。
def leaderboard (self):
""" Return a list of players ranked by score, where score is the total of the count of the players tags and their points. """
return self.players.extra(
select={'score': '(select count(*) from kaos_tag where kaos_tag.tagger_id=kaos_player.id) + points'},
order_by=('-score',)
)
[字段的Django ORDER_BYΣ(可能重複http://stackoverflow.com/questions/3160798/django-order-by-sum-of場) – DrTyrsa
@DTTyrsa我試圖做同樣的事情,但與其中一個字段是一個計數,而不是一個實際的領域。 – borntyping
我會去denolmazition:添加'tags_count'提交給模型,並用信號更新它。或者使用[原始查詢](https://docs.djangoproject.com/zh/dev/topics/db/sql/)作爲替代方案。 – DrTyrsa