您如何使用PostgreSQL将多个结果分组到一个列中，并对每个结果进行计数

如何解决您如何使用PostgreSQL将多个结果分组到一个列中，并对每个结果进行计数

我现在正处于挣扎中，我整天都在找东西，但是我没有设法找出应该做的事情。

我想使用postgres查看我的API（如果需要，可以是一个函数） 这将显示出少数顶级玩家（创造最多赛事的玩家）

我正在寻找的结果将是这样的输出：

class Export < ApplicationRecord
  has_one_attached :item

  def process_pdf!
    pdf_file_name = "#{item.blob.key}_#{item.blob.filename.to_s}"
    xls_file_name = "#{item.blob.key}_#{item.blob.filename.to_s.sub('pdf','')}"
    file = "#{Rails.root}/tmp/#{pdf_file_name}"
    File.open(file,'wb') do |f|
      f.write(pdf.download)
    end

    PdfScraper.call(pdf_file: file,output_name: xls_file_name)
    self.item.attach(
      io: File.open("#{Rails.root}/tmp/#{xls_file_name}"),filename: "#{xls_file_name}.xlsx",content_type: "application/vnd.openxmlformats-officedocument.spreadsheetml.sheet"
    )
    File.delete(file)
    true
  end
end

首先，我有这样的事件表

{"nickname": "Username",...,"total_event_created": 12,"Events":{"eventID": 1,"event_name":CS:go,"total": 6},{"eventID: 2,"event_name": lol,"total": 6} 

我使用CREATE TABLE "event" ( "id" INT GENERATED ALWAYS AS IDENTITY PRIMARY KEY,"user_id" INT NOT NULL REFERENCES "user"("id"),"game_id" INT NOT NULL REFERENCES "game"("id"),"event_time" TIMESTAMPTZ NOT NULL,"duration" INTERVAL,"player_count" INT NOT NULL DEFAULT 1 CONSTRAINT max_player CHECK (player_count <= player_max),"player_max" INT NOT NULL,"description" TEXT,"status" INT NOT NULL DEFAULT 0,"vocal" TEXT); 尝试了很多不同的方法，但是我从来没有设法解决问题。

到目前为止，我最好的尝试是这样：

json_agg()

输出：

SELECT DISTINCT ON (e.user_id) "user_id" AS "_user_id",us.nickname AS "_name",us.avatar AS "_avatar",us.banner AS "_banner",(SELECT COUNT(id) FROM user_access."event" ev WHERE us.id = ev.user_id) AS "_total_events",(SELECT COUNT (CASE WHEN g.id = 1 THEN 1 END) ) AS "_total_cs",(SELECT COUNT (CASE WHEN g.id = 2 THEN 1 END) ) AS "_total_lol"
FROM user_access."user" us
JOIN user_access."event" e  ON us.id = e.user_id 
JOIN user_access."game"  g ON  e.game_id = g.id
GROUP BY  e.user_id,us.nickname,us.id ;

感谢家伙清除了我的代码，从您寄给我的信息来看，我得到了一些更好的信息：

_user_id |   _name    |   _avatar   |   _banner   | _total_events | _total_cs | _total_lol 
----------+------------+-------------+-------------+---------------+-----------+------------
       1 | test1login | avatar1.png | banner1.png |             3 |         2 |          1
       2 | test2login | avatar2.png | banner2.png |             1 |         0 |          1

输出：

SELECT  us."id" AS "_user_id",COUNT(DISTINCT e.id) as "_total_events",jsonb_build_object('cs',COUNT(*) FILTER (WHERE g.id = 1 ),'lol',COUNT(*) FILTER (WHERE g.id = 2))  AS "_total_by_game"       
        FROM user_access."user" us 
        JOIN user_access."event" e ON us.id = e.user_id 
        JOIN user_access."game"  g ON e.game_id = g.id
    GROUP BY us.id
    ORDER BY "_total_events" DESC;

解决方法

您可以使用条件聚合：

SELECT us."user_id" AS "_user_id",us.nickname AS "_name",us.avatar AS "_avatar",us.banner AS "_banner",COUNT(DISTINCT e.id) as "_total_events",COUNT(*) FILTER (WHERE g.id = 1 ) AS "_total_cs",COUNT(*) FILTER (WHERE g.id = 2) AS "_total_lol"
FROM user_access."user" us JOIN
     user_access."event" e
     ON us.id = e.user_id JOIN
     user_access."game"  g
     ON e.game_id = g.id
GROUP BY us.id ;

,

查询中有几个问题：

1. 您按用户分组。您可以进行三次此操作，但没有明显的原因：一次按他们的ID，然后按他们的名字，再按一次ID。
2. 您使用DISTINCT ON用户ID。为什么？ GROUP BY已经为每个用户提供了一行。然后，用DISTINCT ON说：从哪一行给我一行，无论哪一个（因为对于ORDER BY通常没有DISTINCT ON子句，但是，您反正每位用户只有一行，所以这都是无效的。
3. (SELECT COUNT (CASE WHEN g.id = 1 THEN 1 END) )无效，应该引发错误。可以将其重写为CASE WHEN g.id = 1 THEN 1 ELSE 0 END。但是，当您按每个用户进行汇总时，可能会有多个g.id。您要显示哪一个？必须有一个聚合功能才能使其正常工作，例如SUM((SELECT COUNT (CASE WHEN g.id = 1 THEN 1 END)))，当然还是SUM(CASE WHEN g.id = 1 THEN 1 ELSE 0 END)。

似乎您只想进行汇总：

SELECT
  us.id AS "_user_id",COUNT(*) AS "_total_events",COUNT(CASE WHEN g.id = 1 THEN 1 END) AS "_total_cs",COUNT(CASE WHEN g.id = 2 THEN 1 END) AS "_total_lol"
FROM user_access.user us
JOIN user_access.event e ON e.user_id = us.id 
JOIN user_access.game g ON g.id = e.game_id
GROUP BY us.id

对于前5个添加：

ORDER BY COUNT(*) DESC
FETCH FIRST 5 ROWS ONLY; 

顺便说一句：您是否真的想计算用户在游戏1（_total_cs）和游戏2（_total_lol）中发生了多少个事件？还是错别字？