有一个表消息,其中包含如下所示的数据:

Id   Name   Other_Columns
-------------------------
1    A       A_data_1
2    A       A_data_2
3    A       A_data_3
4    B       B_data_1
5    B       B_data_2
6    C       C_data_1

如果我按名称从消息组中运行查询select *,我将得到如下结果:

1    A       A_data_1
4    B       B_data_1
6    C       C_data_1

哪个查询将返回以下结果?

3    A       A_data_3
5    B       B_data_2
6    C       C_data_1

也就是说,应该返回每个组中的最后一条记录。

目前,这是我使用的查询:

SELECT
  *
FROM (SELECT
  *
FROM messages
ORDER BY id DESC) AS x
GROUP BY name

但这看起来效率很低。还有其他方法可以达到同样的效果吗?


当前回答

这是另一个没有子查询的选项。

本解决方案使用MySQL的LAST_VALUE窗口函数,利用窗口函数框架提供MySQL工具。

SELECT DISTINCT 
    LAST_VALUE(Id)            
        OVER(PARTITION BY Name 
             ORDER     BY Id 
             ROWS BETWEEN 0 PRECEDING 
                      AND UNBOUNDED FOLLOWING),
    Name,
    LAST_VALUE(Other_Columns)            
        OVER(PARTITION BY Name 
             ORDER     BY Id 
             ROWS BETWEEN 0 PRECEDING 
                      AND UNBOUNDED FOLLOWING)
FROM   
    tab

在这里试试。

其他回答

SELECT * FROM table_name WHERE primary_key IN (SELECT MAX(primary_key) FROM table_name GROUP BY column_name )

这里有两个建议。首先,如果mysql支持ROW_NUMBER(),这很简单:

WITH Ranked AS (
  SELECT Id, Name, OtherColumns,
    ROW_NUMBER() OVER (
      PARTITION BY Name
      ORDER BY Id DESC
    ) AS rk
  FROM messages
)
  SELECT Id, Name, OtherColumns
  FROM messages
  WHERE rk = 1;

我猜你说的"最后"是指最后一个。如果不是,则相应地更改ROW_NUMBER()窗口的ORDER BY子句。如果ROW_NUMBER()不可用,这是另一个解决方案:

其次,如果没有,这通常是一个很好的方法:

SELECT
  Id, Name, OtherColumns
FROM messages
WHERE NOT EXISTS (
  SELECT * FROM messages as M2
  WHERE M2.Name = messages.Name
  AND M2.Id > messages.Id
)

换句话说,选择没有相同名称的later-Id消息的消息。

我也遇到过类似的问题

子查询和加入救援

SELECT p."Date"
        ,p."Symbol"
        ,p."ratio_roll_qtr_ret"
    FROM PUBLIC."prices_vw" AS p
    JOIN (
        SELECT "Symbol"
            ,max("Date")
        FROM PUBLIC."prices_vw"
        GROUP BY "Symbol"
        ) AS sq ON p."Date" = sq."max"
        AND p."Symbol" = sq."Symbol"
    WHERE p."ratio_roll_qtr_ret" IS NOT NULL
    ORDER BY "ratio_roll_qtr_ret" DESC;

另一种方法:

找到每个程序中最大m2_price的属性(一个程序中有n个属性):

select * from properties p
join (
    select max(m2_price) as max_price 
    from properties 
    group by program_id
) p2 on (p.program_id = p2.program_id)
having p.m2_price = max_price

你可以通过计数来分组,也可以得到分组的最后一项,比如:

SELECT 
    user,
    COUNT(user) AS count,
    MAX(id) as last
FROM request 
GROUP BY user