Question

我有下表：

我需要选择某个日期所有玩家的最高分数。

例如，我需要球员2000-1-1之后取得的最高成绩：预期结果：

我尝试了以下选择：

select *
from player_score ps
where ps.created_at >= '2000-1-1' 
and ps.score = (select max(ps2.score) 
                  from player_score ps2 
                  where ps2.player_id = ps.player_id)

但它只返回一条记录：

您能否告诉我，达到预期效果的最佳方法是什么？谢谢。

Answer 1

由于您希望每位玩家max，请执行group by player_id

select ps.player_id, max(ps.score)
from player_score ps
where ps.created_at >= '2000-1-1' group by ps.player_id

Answer 2

您还需要在子查询中按日期进行过滤：

;with cte (player_id,MaxScore) as (
   select ps.player_id, max(ps.score) as MaxScore
   from player_score ps
   where ps.created_at >= '2000-1-1' 
   group by ps.player_id
)
select ps.*
from player_score ps
inner join cte on cte.player_id = ps.player_id and cte.MaxScore = ps.Score
where ps.created_at >= '2000-1-1'

其他玩家的截止日期之前出现了总的最大值，这就是为什么没有行的原因。

如果您任意地希望每位玩家获得一个记录，即使有最高分的平局，那么请比较ID而不是得分：

select ps.*
from player_score ps
where ps.created_at >= '2000-01-01' 
and ps.score = (select max(ps2.score) 
                from player_score ps2 
                where ps2.created_at >= '2000-01-01' and
                      ps2.player_id = ps.player_id
               );

在Postgres中，最简单的解决方案是：

select ps.*
from player_score ps
where ps.created_at >= '2000-01-01' 
and ps.id = (select ps2.id 
             from player_score ps2 
             where ps2.created_at >= '2000-01-01' and
                   ps2.player_id = ps.player_id
             order by ps2.score desc
             limit 1
            );

Answer 3

除了Gordon Linoff的解决方案之外，您还可以使用IN谓词来解决它。
但是请确保将GROUP BY子句放在子查询中，而不是外部查询中。

SELECT  id, 
        player_id,
        score,
        created_at 
FROM player_score ps2
WHERE ps2.score IN (SELECT MAX(score) FROM player_score ps
            WHERE created_at >= '2000-1-1'  AND ps.player_id = ps2.player_id
            GROUP BY ps.player_id)
ORDER BY player_id

Answer 4

最大值始终是一个值，因此，如果您需要更多值，请联系我们“按订单排序” 它将返回从上到下降序的所有值，并且您可以在表格顶部看到最大值

Answer 5

关于@krithikaGopalakrisnan的答案，假设您需要结果集中的ID和CreatedDate列，则可以使用：

public class CustomValidator : AbstractValidator<Customer> {
  public CustomValidator()
  {
    RuleFor(obj => obj.Prop).NotNull().WithSeverity(Severity.Error);
    RuleFor(obj => obj.Prop).NotEqual("foo").WithSeverity(Severity.Warning);
  }
}

选择某个日期具有最大值的记录

5 个答案: