ActiveRecord为30天的每个日期选择前10个项目

Question

模型是Posts，每个帖子都有upvotes：integer。每天创建数百个帖子，我想通过一个查询返回过去30天内每天排名最高的10个帖子。 （300个结果）

按时间顺序排列

我想返回过去30天的结果，并在同一查询中显示每天的前10条帖子。

预期结果应该是：

[今天的前10个帖子，昨天的前10个帖子... 30天前的前10个帖子。

是否有任何方法可以执行此查询而无需每天进行循环和单独查询？

Answer 1

您可以使用单个查询来获取所有这些查询：

(SELECT * FROM posts WHERE created_at::DATE = CURRENT_DATE ORDER BY upvotes DESC LIMIT 10)
UNION ALL
(SELECT * FROM posts WHERE created_at::DATE = CURRENT_DATE - INTERVAL '1 DAY' ORDER BY upvotes DESC LIMIT 10)
UNION ALL
(SELECT * FROM posts WHERE created_at::DATE = CURRENT_DATE - INTERVAL '30 DAYS' ORDER BY upvotes DESC LIMIT 10)

或三个单个：

Post.where('created_at::date = ?', Date.today).order(upvotes: :desc).limit(10)
Post.where('created_at::date = ?', Date.yesterday).order(upvotes: :desc).limit(10)
Post.where('created_at::date = ?', 30.days.ago).order(upvotes: :desc).limit(10)

Answer 2

1. 原始SQL看起来像这样：

SELECT (array_agg('id => ' || id || ', ' || 'body => ' ||  body || ', ' || 'upvotes => ' || upvotes order by upvotes desc))[1:10] AS posts, date FROM posts GROUP BY date;

它返回具有2个值的散列，第一个是具有包含帖子数组的关键帖子，第二个值是日期

sql = "select (array_agg('id => ' || id || ', ' || 'body => ' ||  body || ', ' || 'upvotes => ' || upvotes order by upvotes desc))[1:
10] as posts, date from posts group by date;"

records = ActiveRecord::Base.connection.execute(sql)
records[0] = {"posts"=> "{\"id => 8, body => atis atis animatis, upvotes => 9\",\"id => 11, body => Fuga enim sit aliquam., upvotes => 8\",\"id => 20, body => Vel vero fugiat explicabo., upvotes => 8\",\"id => 18, body => Nesciunt dolor voluptatibus ad., upvotes => 7\",\"id => 17, body => Laboriosam qui nam esse., upvotes => 6\",\"id => 13, body => Earum est temporibus odio., upvotes => 6\",\"id => 16, body => Autem qui nam omnis., upvotes => 4\",\"id => 7, body => Facilis aut non aut., upvotes => 4\",\"id => 15, body => Voluptatem non asperiores vero., upvotes => 4\",\"id => 12, body => Nihil facere aut natus., upvotes => 3\"}",
 "date"=>2019-11-24 14:41:34 UTC}

然后您可以解析结果。希望对您有所帮助