如何使用“GROUP BY”在单个查询中查找 SQL 中最高和第二高的条目？

Posted 2023-05-09

【中文标题】如何使用“GROUP BY”在单个查询中查找 SQL 中最高和第二高的条目？

【英文标题】：How to find the highest and second highest entry in SQL in a single query using `GROUP BY`?

【发布时间】：2021-12-25 06:34:54

【问题描述】：

假设这是提供的表格。

PID	TID	Type	Freq
1	1	A	3
1	1	A	2
1	1	A	1
1	1	B	3
1	2	A	4
1	2	B	5

我想写一个查询来得到这样的输出。

PID	TID	Type	Max_Freq_1	Max_Freq_2
1	1	A	3	2
1	1	B	3	NULL
1	2	A	4	NULL
1	2	B	5	NULL

也就是说，给定PID、TID、Type 的组合，最高频率和次高频率是多少？如果表中没有足够数量的条目，则将第二高设置为NULL

【参考方案1】：

如果您的数据库可以使用窗口函数，那么可以通过 DENSE_RANK 函数计算前 2 个 Freq。

SELECT PID, TID, Type
, MAX(CASE WHEN Rnk = 1 THEN Freq END) AS Max_Freq_1
, MAX(CASE WHEN Rnk = 2 THEN Freq END) AS Max_Freq_2
FROM
(
    SELECT PID, TID, Type, Freq
    , DENSE_RANK() OVER (PARTITION BY PID, TID, Type ORDER BY Freq DESC) AS Rnk
    FROM YourTable t
) q
GROUP BY PID, TID, Type
ORDER BY PID, TID, Type

pid	tid	type	max_freq_1	max_freq_2
1	1	A	3	2
1	1	B	3	null
1	2	A	4	null
1	2	B	5	null

如果 ROW_NUMBER 不可用，请试试这个。

SELECT PID, TID, Type
, MAX(CASE WHEN Rnk = 1 THEN Freq END) AS Max_Freq_1
, MAX(CASE WHEN Rnk = 2 THEN Freq END) AS Max_Freq_2
FROM
(
    SELECT t1.PID, t1.TID, t1.Type, t1.Freq
    , COUNT(DISTINCT t2.Freq) AS Rnk
    FROM YourTable t1
    LEFT JOIN YourTable t2 
      ON t2.PID = t1.PID
     AND t2.TID = t1.TID
     AND t2.Type = t1.Type
     AND t2.Freq >= t1.Freq
    GROUP BY t1.PID, t1.TID, t1.Type, t1.Freq
) q
GROUP BY PID, TID, Type
ORDER BY PID, TID, Type

dbfiddle here

上的演示

【参考方案2】：

这是我在 PostgreSQL 上提出的。使用 row_number 之类的窗口函数是获得所需结果的最简单方法。

with t as (
    select *, row_number() over (partition by pid, tid, "type" order by freq desc) as r 
    from test_so 
) select pid, tid, "type", max(case when r = 1 then freq end) as "highest", max(case when r = 2 then freq end) as "second_highest" 
    from t 
    group by pid, tid, "type"