加入 2 个表并将重复的行值显示到新列中

Question

我已经花了 2 天的时间来解决这个问题，但找不到任何帮助。

我有 2 张桌子。

1.玩家

id	name
1	AA
2	BB
3	CC
4	DD

2。匹配

id	player_id	match_id	date	has_opponent
1	1	1	2021-06-23	0
2	2	1	2021-06-23	0
3	3	2	2021-06-21	0
4	4	3	2021-06-22	1

预期结果

match_id	date	first_opponent_name	second_opponent_name
1	2021-06-23	AA	BB
2	2021-06-21	CC
3	2021-06-22	DD	DD

我想用新列加入这两个表（ first_opponent_name 和 second_opponent_name ） 如果满足以下条件：

1. 如果有一个重复的 ma​​tch_id 但它有不同的 player_id，则在 2 个单独的列中显示他们的名字。
2. 如果只有 1 个 ma​​tch_id 并且它有 1 个 player_id 并且 has_opponent 列为 0，那么这应该只显示 first_opponent_name
3. 如果只有 1 个 ma​​tch_id 并且它有 1 个 player_id 并且 has_opponent 列为 1，那么 first_opponent_name 和 second_opponent strong> _name 应该显示为相同的值。

Answer 1

未经测试，但我认为这会起作用，基本上它在不同的场景中将表连接到自身，然后检查它匹配的场景。

SELECT
  m.match_id
 ,m.date
 ,p1.name AS first_opponent_name
 ,CASE WHEN p2.name IS NOT NULL THEN p2.name 
       WHEN nop2.match_id IS NOT NULL THEN '' 
       WHEN samep2.match_id IS NOT NULL THEN p1.name END AS second_opponent_name 
FROM
  matching AS m
INNER JOIN players AS p1
  ON m.player_id = p1.player_id 
LEFT JOIN matching AS hasp2
  ON m.match_id = hasp2.match_id
  AND m.player_id <> hasp2.player_id
LEFT JOIN players AS p2
  ON hasp2.player_id = p2.player_id
LEFT JOIN (
  SELECT
    match_id
  FROM
    matching
  WHERE
    has_opponent = 0
  GROUP BY
    match_id
  HAVING COUNT(*) = 1
) AS nop2
  ON m.match_id = nop2.match_id
LEFT JOIN (
  SELECT
    match_id
  FROM
    matching
  WHERE
    has_opponent = 1
) AS samep2
  ON m.match_id = samep2.match_id

可以说，您可以放弃 hasp2 或 samep2 连接，并假设如果不满足其他两种情况，则取决于您对数据的信任程度，以了解这是否是一个可行的选择。

Answer 2

WITH cte AS ( SELECT DISTINCT
                     match_id,
                     `date`,
                     MIN(player_id) OVER (PARTITION BY match_id) p1,
                     MAX(player_id) OVER (PARTITION BY match_id) p2,
                     has_opponent
              FROM matching )
SELECT match_id,
       `date`,
       p1.name first_opponent_name,
       CASE WHEN has_opponent OR (p1 != p2) THEN p2.name END second_opponent_name
FROM cte
JOIN players p1 ON cte.p1 = p1.id
JOIN players p2 ON cte.p2 = p2.id
ORDER BY 1

如果不支持 CTE 和窗口函数，则

SELECT match_id,
       `date`,
       p1.name first_opponent_name,
       CASE WHEN has_opponent OR (p1 != p2) THEN p2.name END second_opponent_name
FROM ( SELECT match_id,
              `date`,
              MIN(player_id) p1,
              MAX(player_id) p2,
              has_opponent
       FROM matching
       GROUP BY match_id,
                `date`,
                has_opponent ) cte
JOIN players p1 ON cte.p1 = p1.id
JOIN players p2 ON cte.p2 = p2.id
ORDER BY 1

https://dbfiddle.uk/?rdbms=mysql_8.0&rdbms2=mysql_5.7&fiddle=4b947a2e7d31d50dfc1242b194b204fe

查询假定相同的match_id 严格匹配相同的date 和match_id。

Answer 3

唯一棘手的部分是处理多个玩家，这非常适合使用窗口函数和row_number()；那么这只是将这些与条件案例聚合起来的案例：

select match_Id, date, 
    Max(case when cnt=1
        then name 
    else
        case when rn=1 then name end
    end) first_opponent_name,

    Max(case when cnt=1 then
        case when has_opponent=1 then name else '' end
    else
        case when rn=2 then name end
    end) second_opponent_name

from (
    select m.match_id, m.date, m.has_opponent, p.name, 
      Row_Number() over(partition by match_id order by player_id)rn, 
      Count(*) over (partition by match_id) cnt
    from matching m
    join players p on p.id=m.player_id
)x
group by match_id, date

见working example Fiddle