Question

如何使用djangos ORM构建单个查询，以便为每位玩家获得低于1.000的最高分？我使用的是django 1.7 rc2和python 3.4。

models.py：

from django.db import models

class Player(models.Model):
    pass

class Score(models.Model):
    player = models.ForeignKey(Player)
    score = models.PositiveIntegerField(default=0)

到目前为止我的解决方案：

tests.py

from django.test import TestCase

from bar.models import Player, Score


class BarTest(TestCase):
    def test_do_it(self):
        p1 = Player.objects.create()
        p2 = Player.objects.create()

        Score.objects.create(player=p1, score=100)
        Score.objects.create(player=p1, score=12000)
        Score.objects.create(player=p2, score=10000)
        Score.objects.create(player=p2, score=500)
        Score.objects.create(player=p2, score=900)

        players = Player.objects.all()
        for player in players:
            score = Score.objects.filter(player=player, score__lte=1000).order_by('score').last()
            if score:
                print(player.id, score.id, score.score)

我不想迭代python中的每个玩家并对每个玩家进行查询，而只用一个查询来解决这个问题（不用编写原始sql）。

Answer 1

使用select_related()，您可以一次性获取所有分数及其玩家：

from collections import defaultdict
scores = defaultdict(list)

# Fetch all scores (with score <= 1000), including their players.
for score in Score.objects.filter(score__lte=1000).select_related('player'):
    # Group scores by their player.
    scores[score.player.id].append(score)

# Get max score for each player.
for p in scores:
    max_score = max(scores[p], key=lambda s: s.score)
    print(p, max_score.id, max_score.score)

Answer 2

怎么样：

all_scores = Score.objects.filter(score__lte=1000).order_by('player', '-score')
current_player = None
scores = []
for score in all_scores:
    if current_player != score.player:
        current_player = score.player
        scores.append(score)

只有一次迭代超过分数。排序由数据库完成。

修改

最后，我找到了一个解决方案：

Player.objects.filter(score__score__lte=1000 ).annotate(max_score=Max('score__score'))

构建此SQL查询：

SELECT "bar_player"."id", MAX("bar_score"."score") AS "max_score" FROM "bar_player" LEFT OUTER JOIN "bar_score" ON ( "bar_player"."id" = "bar_score"."player_id" ) WHERE "bar_score"."score" <= 1000 GROUP BY "bar_player"."id"

如何将多个查询减少到只有一个查询？

2 个答案: