SQL JOIN 只返回第一个匹配项

Question

我认为这是一个基本问题，但我无法弄清楚。我是新手，所以请多多包涵。

我正在分析 FIFA 比赛中的球员数据，我想获得一个表格，其中包含每个球员的最高评分以及发生这种情况的最早年龄。

这是数据示例：

id	name	position	rating	age
1	James	RW	70	20
1	James	RW	71	21
2	Frank	CB	73	23
2	Frank	CB	73	24
3	Miles	CM	75	27
3	Miles	CM	74	28

这是查询应该返回的内容：

id	name	position	rating	age
1	James	RW	71	21
2	Frank	CB	73	23
3	Miles	CM	75	27

我以为我可以先让每个球员的总分最高，然后通过JOIN 得到年龄，但这样其他年份的球员也有相同的最高评分。

id	name	position	rating	age
1	James	RW	71	21
2	Frank	CB	73	23
2	Frank	CB	73	24
3	Miles	CM	75	27

谢谢，

Answer 1

一种方法使用ROW_NUMBER 和QUALIFY：

SELECT *
FROM yourTable
WHERE true
QUALIFY ROW_NUMBER() OVER (PARTITION BY id ORDER BY rating DESC, age) = 1;

Answer 2

试试这个（可能会快一点）：

with mytable as (
    select 1 as id, "James" as name, "RW" as position, 70 as rating,  20 as age union all 
    select 1, "James", "RW", 71, 21 union all
    select 2, "Frank", "CB", 73, 23 union all
    select 2, "Frank", "CB", 73, 24 union all
    select 3, "Miles", "CM", 75, 27 union all
    select 3, "Miles", "CM", 74, 28
)
select array_agg(t order by rating desc, age asc limit 1)[OFFSET(0)].*,
from mytable as t
group by t.id

Answer 3

如果你愿意，也可以试试这个。我认为这是您需要的。

select p1.id,
       p1.name,
       p1.pos,
       max(p1.rating),
       p1.age
  from players p1
  join (select name, min(age) min_age
          from players 
         group by 1) p2 on p1.name = p2.name and p1.age = p2.min_age
 group by p1.id, p1.name, p1.pos, p1.age