MySQL Query для всего

Question

У меня есть три таблицы

Я выполнил этот запрос:

SELECT 
`tbl_players`.`id` AS `player_id`
,`tbl_players`.`player_name`
,`tbl_sports`.`sports_name`
,`tbl_scores`.`score`
FROM `tbl_players`
LEFT JOIN `tbl_scores` ON `tbl_players`.`id`=`tbl_scores`.`player_id`
LEFT JOIN `tbl_sports` ON `tbl_players`.`sports_id`=`tbl_sports`.`sports_id`
ORDER BY `tbl_players`.`id` ASC

И я получил это:

Теперь я хочу это:

Answer 1

Это может помочь.

Вам нужно PIVOT свои данные

SELECT 
 player_id,
 player_name,
 sports_name,
 sum(case when tbl_scores.season=1 then `tbl_scores`.`score` end) as season1score,
 sum(case when tbl_scores.season=2 then `tbl_scores`.`score` end) as season2score,
 sum(case when tbl_scores.season=3 then `tbl_scores`.`score` end) as season3score
from 
  tbl_players
left join  tbl_sports 
  on `tbl_players`.`id`=`tbl_scores`.`player_id`
left join sports 
  on `tbl_players`.`sports_id`=`tbl_sports`.`sports_id`
group by
 layer_id,
 player_name,
 sports_name,

Answer 2

использовать условное агрегирование

 SELECT 
    `tbl_players`.`id` AS `player_id`
    ,`tbl_players`.`player_name`
    ,`tbl_sports`.`sports_name`
    , sum(case when tbl_scores.season=1 then `tbl_scores`.`score` end) as season1score
   ,sum(case when tbl_scores.season=2 then `tbl_scores`.`score` end) as season2score
,sum(case when tbl_scores.season=3 then `tbl_scores`.`score` end) as season3score
    FROM `tbl_players`
    LEFT JOIN `tbl_scores` ON `tbl_players`.`id`=`tbl_scores`.`player_id`
    LEFT JOIN `tbl_sports` ON `tbl_players`.`sports_id`=`tbl_sports`.`sports_id`
group by `tbl_players`.`player_name`
    ,`tbl_sports`.`sports_name`,player_id

Answer 3

Я считаю, что вы должны использовать внутреннее соединение вместо левого соединения

Answer 4

использовать условное агрегирование

select player_id,player_name,sports_name,
       sum(case when season=1 then score end) as season1_score,
       sum(case when season=2 then score end) as season2_score,
       sum(case when season=3 then score end) as season3_score
from
(
SELECT 
`tbl_players`.`id` AS `player_id`
,`tbl_players`.`player_name`
,`tbl_sports`.`sports_name`
,`tbl_scores`.`score`,`tbl_scores`.`season`
FROM `tbl_players`
LEFT JOIN `tbl_scores` ON `tbl_players`.`id`=`tbl_scores`.`player_id`
LEFT JOIN `tbl_sports` ON `tbl_players`.`sports_id`=`tbl_sports`.`sports_id`
)A group by player_id,player_name,sports_name

Answer 5

Используйте «отличный» выбор, чтобы не получать дубликаты.

Просто добавьте «DISTINCT» сразу после «SELECT» в вашем запросе.

Пример: SELECT DISTINCT column1, column2, ... FROM table_name;