他のテーブルの2つのCOUNT（）に条件がある場合のクエリ

Question

スポーツのリストを含むスポーツと呼ばれるテーブル、特定のスポーツの季節を含む季節と競争のある競技特定のスポーツと季節。

スポーツのリストを印刷するには、季節や競争の度合いをそれぞれ1つのMySQLクエリで確認する必要があります。私のテーブル構造：

スポーツ

+--------------------+------------------+------+-----+---------+----------------+ 
| Field    | Type    | Null | Key | Default | Extra   | 
+--------------------+------------------+------+-----+---------+----------------+ 
| id     | int(10) unsigned | NO | PRI | NULL | auto_increment | 
| name    | varchar(32)  | NO |  | NULL |    | 
| slug    | varchar(45)  | NO |  | NULL |    | 
| description  | varchar(128)  | NO |  | NULL |    | 
+--------------------+------------------+------+-----+---------+----------------+ 

季節

+--------------------+------------------+------+-----+---------+----------------+ 
| Field    | Type    | Null | Key | Default | Extra   | 
+--------------------+------------------+------+-----+---------+----------------+ 
| id     | int(10) unsigned | NO | PRI | NULL | auto_increment | 
| id_sport   | int(10) unsigned | NO | MUL | NULL |    | 
| name    | varchar(32)  | NO |  | NULL |    | 
| slug    | varchar(32)  | NO |  | NULL |    | 
+--------------------+------------------+------+-----+---------+----------------+ 

競技

+--------------------+------------------+------+-----+---------+----------------+ 
| Field    | Type    | Null | Key | Default | Extra   | 
+--------------------+------------------+------+-----+---------+----------------+ 
| id     | int(10) unsigned | NO | PRI | NULL | auto_increment | 
| id_season   | int(10) unsigned | NO | MUL | NULL |    | 
| name    | varchar(32)  | NO |  | NULL |    | 
| slug    | varchar(64)  | NO |  | NULL |    | 
| description  | varchar(128)  | YES |  | NULL |    | 
+--------------------+------------------+------+-----+---------+----------------+ 

クエリの結果には、スポーツ。*、total_seasons（seasons.id_sport = sports.idのシーズンの合計）とtotal_competitions（competition.id_season = seasons.idとseasons.id_sport = sports.idの競合の合計）が含まれている必要があります。 。

ありがとうございます！

Answer 1

使用は

select 
    sports.id, 
    sports.name, 
    sports.slug, 
    sports.description, 
    ifnull(count(distinct seasons.id), 0) as DistinctSeasons, 
    ifnull(count(distinct competitions.id), 0) as TotalCompetitions 
from 
    sports 
    left join seasons on sports.id=seasons.id_sport 
    left join competitions on seasons.id = competitions.id_season 
group by 
    sports.id; 

Answer 2

SELECT sports.*, 
    COUNT(seasons.id) AS total_seasons, 
    COUNT(competitions.id) AS total_competitions 
    FROM sports, seasons, competitions 
    WHERE sports.id=id_sport AND seasons.id=id_season 
    GROUP BY sports.id 

Answer 3

select 
     sports.id, 
     sports.name, 
     sports.slug, 
     sports.description, 
     count(distinct seasons.id) as DistinctSeasons, 
     count(*) as TotalCompetitions 
    from 
     sports 
     left outer join Seasons 
      on sports.id = seasons.id_sport, 
     left outer join Competitions 
      on seasons.id = Competitions.id_season 
    group by 
     sports.id