查找最高值列各组

Question

我有一个元数据表看起来像这样的：

+----------------------------+ 
|  metadata   | 
+----+----------+------------+ 
| id | group_id | message_id | 
+----+----------+------------+ 
| 1 | 1  |  5  | 
+----+----------+------------+ 
| 2 | 2  |  6  | 
+----+----------+------------+ 
| 3 | 1  |  8  | 
+----+----------+------------+ 
| 4 | 2  |  10  | 
+----+----------+------------+ 

对于每个group_id，我想选择具有最高id 行所以我想回来：

+----------------------------+ 
|   results   | 
+----+----------+------------+ 
| id | group_id | message_id | 
+----+----------+------------+ 
| 3 | 1  |  8  | 
+----+----------+------------+ 
| 4 | 2  |  10  | 
+----+----------+------------+ 

我在想这将是这样的事情，但以下只返回行4。从我想选择MAX(metadata.id)

SELECT 
    MAX(metadata.id) 
FROM 
    metadata 
GROUP BY 
    metadata.message_id 

假设Metadatum是一个ActiveRecord模型组，每组，并group是belongs_to关系，你将如何构建这个查询与AR？

Answer 1

产生的预期结果查询（假设id是唯一的）：

SELECT t1.* FROM t AS t1 
JOIN (
    SELECT max(id) AS id FROM t 
    GROUP BY group_id 
) t2 ON t1.id = t2.id 

输出：

| ID | GROUP_ID | MESSAGE_ID | 
|----|----------|------------| 
| 3 |  1 |   8 | 
| 4 |  2 |   10 | 

小提琴here。

但是，我不完全明白你的意思是由AR查询。要做到这一点

Answer 2

一种方法是使用row_number()：

SELECT m.id 
FROM (select m.*, 
      row_number() over (partition by m.message_id order by m.id desc) as seqnum 
     from metadata m 
    ) m 
WHERE seqnum = 1; 

在Postgres里，你也可以用distinct on做到这一点：

select distinct on (m.message_id), m.id 
from metadata m 
order by m.message_id, m.id; 

Answer 3

正如编程的乐趣，这里是另一种解决方案

SELECT t1.* 
FROM metadata AS t1 
WHERE t1.id = (
    SELECT t2.id 
    FROM metadata AS t2 
    WHERE t2.group_id = t1.group_id 
    ORDER BY t2.id DESC 
    LIMIT 1 
) 

Answer 4

SELECT DISTINCT ON (group_id) *  
FROM metadata 
ORDER BY group_id, id DESC; 

-> SQLfiddle

就像：
Select first row in each GROUP BY group?