Question

我有两个表格charts和chart_entries，我希望获得当前前5个图表位置（基于在给定位置选择最新的chart_entries）。

有没有办法让这个单一的查询？理想情况下，我的图表将有多达500个位置，因此这样的多个查询实际上是不可行的。

在rails中，我目前正在Chart.rb

中使用此方法

def table
  (1..5).each do |i|
    chart_entries.where(position: i).order('entered_at DESC').first
  end
end

生成以下SQL

SELECT "chart_entries".* FROM "chart_entries" WHERE "chart_entries"."chart_id" = 1 AND "chart_entries"."position" = 1 ORDER BY entered_at DESC LIMIT 1
SELECT "chart_entries".* FROM "chart_entries" WHERE "chart_entries"."chart_id" = 1 AND "chart_entries"."position" = 2 ORDER BY entered_at DESC LIMIT 1
SELECT "chart_entries".* FROM "chart_entries" WHERE "chart_entries"."chart_id" = 1 AND "chart_entries"."position" = 3 ORDER BY entered_at DESC LIMIT 1
SELECT "chart_entries".* FROM "chart_entries" WHERE "chart_entries"."chart_id" = 1 AND "chart_entries"."position" = 4 ORDER BY entered_at DESC LIMIT 1
SELECT "chart_entries".* FROM "chart_entries" WHERE "chart_entries"."chart_id" = 1 AND "chart_entries"."position" = 5 ORDER BY entered_at DESC LIMIT 1

这是表结构（postgresql）

CREATE TABLE chart_entries (
  id integer NOT NULL,
  chart_id integer,
  entity_id integer,
  "position" integer,
  entered_at timestamp without time zone,
  locale character varying(255)
);

CREATE TABLE charts (
  id integer NOT NULL,
  name character varying(255)
);

这是我的最终解决方案

chart_entries.where(position: [1..5]).group('position, id').having('entered_at = MAX(entered_at)').order('position ASC')

Answer 1

这对GROUP BY来说是个很好的选择：

SELECT
    *
FROM
    chart_entries
WHERE
    chart_id = 1
    AND position BETWEEN 1 AND 5
GROUP BY
    position
HAVING
    entered_at = MAX(entered_at)

使用GROUP BY将为每个位置选择一行。 HAVING子句应该按日期选择最新的子句（您也可以使用主键，因为应该是最新的 - 除非您插入自定义日期）。

您可以在position子句中使用BETWEEN来获取 1和5之间的所有位置，而不是需要编写每个WHERE个可用内容。（或者你的范围是什么）。

Answer 2

使用MAX＆amp; GROUP BY

SELECT MAX("chart_entries".id) FROM "chart_entries" 
WHERE "chart_entries"."chart_id" = 1 AND "chart_entries"."position" IN (1,2,3,4,5) 
ORDER BY entered_at DESC 
GROUP BY  "chart_entries"."position"

根据可变条件选择多个单个（LIMIT 1）记录

2 个答案: