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

试试这个:

SELECT jos_categories.title AS name,
       joined .catid,
       joined .title,
       joined .introtext
FROM   jos_categories
       INNER JOIN (SELECT *
                   FROM   (SELECT `title`,
                                  catid,
                                  `created`,
                                  introtext
                           FROM   `jos_content`
                           WHERE  `sectionid` = 6
                           ORDER  BY `id` DESC) AS yes
                   GROUP  BY `yes`.`catid` DESC
                   ORDER  BY `yes`.`created` DESC) AS joined
         ON( joined.catid = jos_categories.id )  

其他回答

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

如果您需要分组查询中文本列的最新或最古老的记录,并且不希望使用子查询,您可以这样做…

例如,你有一个电影列表,需要获得系列电影和最新电影的数量

id series name
1 Star Wars A New hope
2 Star Wars The Empire Strikes Back
3 Star Wars Return of The Jedi
SELECT COUNT(id), series, SUBSTRING(MAX(CONCAT(id, name)), LENGTH(id) + 1), 
FROM Movies
GROUP BY series

这将返回……

id series name
3 Star Wars Return of The Jedi

MAX将返回值最高的行,因此通过将id连接到名称,您现在将获得最新的记录,然后去掉id以获得最终结果。

比使用子查询更有效。

对于给定的例子:

SELECT MAX(Id), Name, SUBSTRING(MAX(CONCAT(Id, Other_Columns)), LENGTH(Id) + 1), 
FROM messages
GROUP BY Name

快乐编码,“愿原力与你同在”:)

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

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。

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