MySQLのクエリが非常に遅い（Left-JOIN、GROUP BYなど？）

Question

MySQLサーバに提出する後続のクエリに問題があります。 25秒かかる...（COUNT（*）< 20k） - 表記は600k行しかありません。しかし、索引は、それがどこにあるべきか（つまり、ON句の関係する列に対して）作成されます。私はGROUP BYを削除しようとしましたが、これはケースを少し改善しました。しかし、クエリは依然として遅い応答に一般的なルールを与えます。私はstackoverflowに見つかったさまざまなケースの解決策を見つけることができなかったので、その投稿をしました。なにか提案を？

SELECT 
    doctor.id as doctor_id, 
    doctor.uuid as doctor_uuid, 
    doctor.firstname as doctor_firstname, 
    doctor.lastname as doctor_lastname, 
    doctor.cloudRdvMask as doctor_cloudRdvMask, 

    GROUP_CONCAT(recommendation.id SEPARATOR ' ') as recommendation_ids, 
    GROUP_CONCAT(recommendation.uuid SEPARATOR ' ') as recommendation_uuids, 
    GROUP_CONCAT(recommendation.disponibility SEPARATOR ' ') as recommendation_disponibilities, 
    GROUP_CONCAT(recommendation.user_id SEPARATOR ' ') as recommendation_user_ids, 
    GROUP_CONCAT(recommendation.user_uuid SEPARATOR ' ') as recommendation_user_uuids, 

    location.id as location_id, 
    location.uuid as location_uuid, 
    location.lat as location_lat, 
    location.lng as location_lng, 

    profession.id as profession_id, 
    profession.uuid as profession_uuid, 
    profession.name as profession_name 
FROM featured as doctor 
LEFT JOIN location as location 
    ON doctor.location_id = location.id 
LEFT JOIN profession as profession 
    ON doctor.profession_id = profession.id 
LEFT JOIN 
    (
     SELECT 
      featured.id as id, 
      featured.uuid as uuid, 
      featured.doctor_id as doctor_id, 
      featured.disponibility as disponibility, 

      user.id as user_id, 
      user.uuid as user_uuid 
     FROM featured as featured 
      LEFT JOIN user as user 
       ON featured.user_id = user.id 
      WHERE discr = 'recommendation' 
     ) as recommendation 
     ON recommendation.doctor_id = doctor.id 
WHERE 
    doctor.discr = 'doctor' 
    AND 
    doctor.state = 'Publié' 
GROUP BY doctor.uuid 

はここで、EXPLAIN結果が来る：

id | select_type | table  | partitions | type | possible_keys    | key     | key_len | ref       | rows | filtered | Extra  | 

1 | SIMPLE  | doctor  | NULL  | ref | discr,state    | discr    | 767  | const       | 194653 | 50.00 | Using where | 
1 | SIMPLE  | location | NULL  | eq_ref | PRIMARY     | PRIMARY    | 4  | doctoome.doctor.location_id |  1 | 100.00 | NULL  | 
1 | SIMPLE  | profession | NULL  | eq_ref | PRIMARY     | PRIMARY    | 4  | doctoome.doctor.profession_id |  1 | 100.00 | NULL  | 
1 | SIMPLE  | featured | NULL  | ref | IDX_3C1359D487F4FB17,discr | IDX_3C1359D487F4FB17 | 5  | doctoome.doctor.id   | 196 | 100.00 | Using where | 
1 | SIMPLE  | user  | NULL  | eq_ref | PRIMARY     | PRIMARY    | 4  | doctoome.featured.user_id  |  1 | 100.00 | Using index | 

EDITこのリンクは、私を助け、それが8秒で今行きます。 https://www.percona.com/blog/2016/10/12/mysql-5-7-performance-tuning-immediately-after-installation/。しかし、私はまだそれが遅いと思う、私はちょうど誰もが改善される可能性があることを知っている場合はそれを聞かせてください。そして、あなたはfeatured(discr, state, doctor_id, location_id, profession_id)とfeatured(doctor_id, discr, user_id)にインデックスをしたい

SELECT . . . -- you need to fix the `GROUP_CONCAT()` column references 
FROM featured doctor LEFT JOIN 
    location location 
    ON doctor.location_id = location.id LEFT JOIN 
    profession profession 
    ON doctor.profession_id = profession.id LEFT JOIN 
    featured featured 
    ON featured.doctor_id = doctor.doctor_id LEFT JOIN 
    user user 
    ON featured.user_id = user.id 
WHERE doctor.discr = 'doctor' AND 
     doctor.state = 'Publié' AND 
     featured.discr = 'recommendation' 
GROUP BY doctor.uuid; 

を：ありがとう

Answer 1

私はサブクエリを削除すると、いくつかの複数のインデックスと一緒に、役立つかもしれないと思います。