23

我正在尝试计算玩家每周玩的次数,如下所示:

player.game_objects.extra(
    select={'week': 'WEEK(`games_game`.`date`)'}
).aggregate(count=Count('week'))

但是 Django 抱怨说

FieldError: Cannot resolve keyword 'week' into field. Choices are: <lists model fields>

我可以像这样在原始 SQL 中做到这一点

SELECT WEEK(date) as week, COUNT(WEEK(date)) as count FROM games_game
WHERE player_id = 3
GROUP BY week

有没有在 Django 中执行原始 SQL 的好方法?

4

2 回答 2

15

您可以使用自定义聚合函数来生成查询:

WEEK_FUNC = 'STRFTIME("%%%%W", %s)' # use 'WEEK(%s)' for mysql

class WeekCountAggregate(models.sql.aggregates.Aggregate):
    is_ordinal = True
    sql_function = 'WEEK' # unused
    sql_template = "COUNT(%s)" % (WEEK_FUNC.replace('%%', '%%%%') % '%(field)s')

class WeekCount(models.aggregates.Aggregate):
    name = 'Week'
    def add_to_query(self, query, alias, col, source, is_summary):
        query.aggregates[alias] = WeekCountAggregate(col, source=source, 
            is_summary=is_summary, **self.extra)


>>> game_objects.extra(select={'week': WEEK_FUNC % '"games_game"."date"'}).values('week').annotate(count=WeekCount('pk'))

但是由于这个 API 没有记录并且已经需要一些原始 SQL,所以最好使用原始查询

于 2011-03-04T15:13:00.743 回答
4

这是问题的示例和不理想的解决方法。以这个示例模型为例:

class Rating(models.Model):
    RATING_CHOICES = (
        (1, '1'),
        (2, '2'),
        (3, '3'),
        (4, '4'),
        (5, '5'),
    )
    rating = models.PositiveIntegerField(choices=RATING_CHOICES)
    rater = models.ForeignKey('User', related_name='ratings_given')
    ratee = models.ForeignKey('User', related_name='ratings_received')

此示例聚合查询以与您相同的方式失败,因为它尝试引用使用创建的非字段值.extra()

User.ratings_received.extra(
    select={'percent_positive': 'ratings > 3'}
).aggregate(count=Avg('positive'))

一种解决方法

可以通过在额外值的定义中使用聚合数据库函数(在本例中为 Avg)直接找到所需的值:

User.ratings.extra(
    select={'percent_positive': 'AVG(rating >= 3)'}
)

此查询将生成以下 SQL 查询:

SELECT (AVG(rating >= 3)) AS `percent_positive`,
       `ratings_rating`.`id`,
       `ratings_rating`.`rating`,
       `ratings_rating`.`rater_id`,
       `ratings_rating`.`ratee_id`
FROM `ratings_rating`
WHERE `ratings_rating`.`ratee_id` = 1

尽管此查询中有不需要的列,我们仍然可以通过隔离值从中获取所需的percent_positive值:

User.ratings.extra(
    select={'percent_positive': 'AVG(rating >= 3)'}
).values('percent_positive')[0]['percent_positive']
于 2011-03-03T17:42:59.980 回答