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 )

如果需要每个Name的最后一行,那么可以按Name为每个行组提供行号,并按Id降序排序。

查询

SELECT t1.Id, 
       t1.Name, 
       t1.Other_Columns
FROM 
(
     SELECT Id, 
            Name, 
            Other_Columns,
    (
        CASE Name WHEN @curA 
        THEN @curRow := @curRow + 1 
        ELSE @curRow := 1 AND @curA := Name END 
    ) + 1 AS rn 
    FROM messages t, 
    (SELECT @curRow := 0, @curA := '') r 
    ORDER BY Name,Id DESC 
)t1
WHERE t1.rn = 1
ORDER BY t1.Id;

SQL小提琴

希望以下Oracle查询能有所帮助:

WITH Temp_table AS
(
    Select id, name, othercolumns, ROW_NUMBER() over (PARTITION BY name ORDER BY ID 
    desc)as rank from messages
)
Select id, name,othercolumns from Temp_table where rank=1

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;

解决方案由子查询小提琴链接

select * from messages where id in
(select max(id) from messages group by Name)

解决方案通过连接条件小提琴链接

select m1.* from messages m1 
left outer join messages m2 
on ( m1.id<m2.id and m1.name=m2.name )
where m2.id is null

这篇文章的原因是只给小提琴链接。 在其他答案中已经提供了相同的SQL。