可能会覆盖计数的工作方式,或者找到更好的方法,完全做到这一点

发布于 2024-12-07 20:37:24 字数 2075 浏览 0 评论 0原文

我的艺术家模型中有一个范围,它按照艺术家在特定时间段内的受欢迎程度排列。 popularity_caches 表中的popularity 每天都会计算一次。

scope :by_popularity, lambda { |*args|
  options = (default_popularity_options).merge(args[0] || {})

  select("SUM(popularity) AS popularity, artists.*").
from("popularity_caches FORCE INDEX (popularity_cache_group), artists FORCE INDEX (index_artists_on_id_and_genre_id)").
where("popularity_caches.target_type = 'Artist'").
where("popularity_caches.target_id = artists.id").
where("popularity_caches.time_frame = ?", options[:time_frame]).
where("popularity_caches.started_on > ?", options[:started_on]).
where("popularity_caches.started_on < ?", options[:ended_on]).
group("artists.id").
order("popularity DESC")
}

这似乎有效,除非我想获取计数:Artist.by_popularity.count。我得到一个时髦的哈希值作为回报(可能是该时期内拥有流行度缓存的艺术家的数量):

#<OrderedHash {295954=>1, 20143=>1, 157532=>1, 181291=>1, 300086=>1, 50100=>1, 262898=>1, 293888=>1, 130158=>2, 279943=>1, 336758=>1, 100201=>1, 134290=>2, 22726=>3, 144620=>2, 62497=>2 # snip

这是我可能想要的 SQL 回报:

SELECT COUNT(DISTINCT(artists.id)) AS count_all
FROM popularity_caches FORCE INDEX (popularity_cache_group), artists FORCE INDEX (index_artists_on_id_and_genre_id)
WHERE (popularity_caches.target_type = 'Artist')
  AND (popularity_caches.target_id = artists.id)
  AND (popularity_caches.time_frame = 'week')
  AND (popularity_caches.started_on > '2011-02-28 16:00:00')
  AND (popularity_caches.started_on < '2011-10-05')
ORDER BY popularity DESC

为了获得计数,我必须创建一个单独的方法,它几乎可以做同样的事情,只不过 SQL 的构成不同。它有点糟糕,因为当我想要分页时,我必须传递两件事:

@artists = Artists.by_popularity(some args).paginate(
  :total_entries => Artist.count_by_popularity(pass in the same args here as in Artist.by_popularity),
  :per_page => 5,
  page => ...
)

这对我来说很臭,因为它非常脆弱。

ARel 有办法做到这一点吗?也许会覆盖它对事物的计数方式(不同的artists.id)并删除group by,这样它就不会返回计数的哈希值?

谢谢!

I have this scope in my artist model that gives me the artists, in the order of their popularity within a certain time period. popularity in the popularity_caches table is computed every day.

scope :by_popularity, lambda { |*args|
  options = (default_popularity_options).merge(args[0] || {})

  select("SUM(popularity) AS popularity, artists.*").
from("popularity_caches FORCE INDEX (popularity_cache_group), artists FORCE INDEX (index_artists_on_id_and_genre_id)").
where("popularity_caches.target_type = 'Artist'").
where("popularity_caches.target_id = artists.id").
where("popularity_caches.time_frame = ?", options[:time_frame]).
where("popularity_caches.started_on > ?", options[:started_on]).
where("popularity_caches.started_on < ?", options[:ended_on]).
group("artists.id").
order("popularity DESC")
}

This seems to work except when I want to get the count: Artist.by_popularity.count. I get a funky hash in return (probably the count of artists that have popularity_caches within that period):

#<OrderedHash {295954=>1, 20143=>1, 157532=>1, 181291=>1, 300086=>1, 50100=>1, 262898=>1, 293888=>1, 130158=>2, 279943=>1, 336758=>1, 100201=>1, 134290=>2, 22726=>3, 144620=>2, 62497=>2 # snip

This is the SQL I probably want in return:

SELECT COUNT(DISTINCT(artists.id)) AS count_all
FROM popularity_caches FORCE INDEX (popularity_cache_group), artists FORCE INDEX (index_artists_on_id_and_genre_id)
WHERE (popularity_caches.target_type = 'Artist')
  AND (popularity_caches.target_id = artists.id)
  AND (popularity_caches.time_frame = 'week')
  AND (popularity_caches.started_on > '2011-02-28 16:00:00')
  AND (popularity_caches.started_on < '2011-10-05')
ORDER BY popularity DESC

To get the count, I had to make a separate method that pretty much does the same thing, except the SQL is formed differently. It kinds sucks through, because when I want to paginate, I have to pass two things:

@artists = Artists.by_popularity(some args).paginate(
  :total_entries => Artist.count_by_popularity(pass in the same args here as in Artist.by_popularity),
  :per_page => 5,
  page => ...
)

That smells to me because it's very brittle.

Is there a way to do this in ARel? Maybe override how it counts things (distinct artists.id) and removing the group by so it doesn't return a hash for the count?

Thanks!

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梦醒灬来后我 2024-12-14 20:37:24

通过令人惊叹的 scuttle.io 解决了:

PopularityCach.select(
  Arel::Nodes::Group.new(Artist.arel_table[:id]).count.as('count_all')
).where(
  PopularityCach.arel_table[:target_type].eq('Artist').and(
    PopularityCach.arel_table[:target_id].eq(Artist.arel_table[:id]).and(
      PopularityCach.arel_table[:time_frame].eq('week').and(
        PopularityCach.arel_table[:started_on].gt('2011-02-28 16:00:00').and(
          PopularityCach.arel_table[:started_on].lt('2011-10-05')
        )
      )
    )
  )
).order(:popularity).reverse_order

Solved with the amazing scuttle.io:

PopularityCach.select(
  Arel::Nodes::Group.new(Artist.arel_table[:id]).count.as('count_all')
).where(
  PopularityCach.arel_table[:target_type].eq('Artist').and(
    PopularityCach.arel_table[:target_id].eq(Artist.arel_table[:id]).and(
      PopularityCach.arel_table[:time_frame].eq('week').and(
        PopularityCach.arel_table[:started_on].gt('2011-02-28 16:00:00').and(
          PopularityCach.arel_table[:started_on].lt('2011-10-05')
        )
      )
    )
  )
).order(:popularity).reverse_order
~没有更多了~
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