Sql 按最大日期和id分组

我有下表：

 id         value       date
 ---------  ---------  ----------
 1          324          2017-05-19 02:00:00
 2          564          2017-05-19 04:00:00
 3          675          2017-05-19 03:00:00
 3          642          2017-05-19 05:00:00
 4          642          2017-05-19 07:00:00
 5          642          2017-05-19 06:00:00
 6          642          2017-05-19 05:00:00
 6          642          2017-05-19 12:00:00
 7          642          2017-05-19 01:00:00
 8          642          2017-05-19 13:00:00
 8          642          2017-05-19 02:00:00

我想按

id

列对

进行分组，但在按
date
降序排序时，还要从第一行选择
value
和
date
列，其中
id

结果应该是这样的：

 id         value       date
 ---------  ---------  ----------
 1          324          2017-05-19 02:00:00
 2          564          2017-05-19 04:00:00
 3          642          2017-05-19 05:00:00
 4          642          2017-05-19 07:00:00
 5          642          2017-05-19 06:00:00
 6          642          2017-05-19 12:00:00
 7          642          2017-05-19 01:00:00
 8          642          2017-05-19 13:00:00

试试这个

select id, value, date 
from  name_table
group by id, value, date 

select id, value, max(date) 
from table_name 
group by id,value; 

试试这个

select id, value, date 
from  name_table
group by id, value, date 

select id, value, max(date) 
from table_name 
group by id,value; 

试试这个

select id, value, date 
from  name_table
group by id, value, date 

select id, value, max(date) 
from table_name 
group by id,value; 

试试这个

select id, value, date 
from  name_table
group by id, value, date 

select id, value, max(date) 
from table_name 
group by id,value; 

像这样的方法应该会奏效：

SELECT t1.id, t1.value, t1.date
FROM your_table t1
INNER JOIN (
    SELECT id, MAX(date) date
    FROM your_table
    GROUP BY id
) t2
ON t1.id = t2.id AND t1.date = t2.date

像这样的方法应该会奏效：

SELECT t1.id, t1.value, t1.date
FROM your_table t1
INNER JOIN (
    SELECT id, MAX(date) date
    FROM your_table
    GROUP BY id
) t2
ON t1.id = t2.id AND t1.date = t2.date

请尝试此查询

SELECT t1.id,value,t1.date FROM new_table t1   
JOIN 
(SELECT nt.id,max(nt.date) AS date  FROM new_table as nt group by nt.id ) AS t2 
ON t1.date = t2.date AND t1.id = t2.id;

它会起作用的

如果有任何疑问，请询问。
尝试此查询

SELECT t1.id,value,t1.date FROM new_table t1   
JOIN 
(SELECT nt.id,max(nt.date) AS date  FROM new_table as nt group by nt.id ) AS t2 
ON t1.date = t2.date AND t1.id = t2.id;

;With cte( id, value ,date)
AS
(

 SELECT 1, 324,'2017-05-19 02:00:00' Union all
 SELECT 2, 564,'2017-05-19 04:00:00' Union all
 SELECT 3, 675,'2017-05-19 03:00:00' Union all
 SELECT 3, 642,'2017-05-19 05:00:00' Union all
 SELECT 4, 642,'2017-05-19 07:00:00' Union all
 SELECT 5, 642,'2017-05-19 06:00:00' Union all
 SELECT 6, 642,'2017-05-19 05:00:00' Union all
 SELECT 6, 642,'2017-05-19 12:00:00' Union all
 SELECT 7, 642,'2017-05-19 01:00:00' Union all
 SELECT 8, 642,'2017-05-19 13:00:00' Union all
 SELECT 8, 642,'2017-05-19 02:00:00' 
 )
SELECT id
    ,value
    ,DATE
FROM (
    SELECT *
        ,ROW_NUMBER() OVER (
            PARTITION BY id ORDER BY id
            ) RNO
    FROM (
        SELECT id
            ,value
            ,MAX([date]) OVER (
                PARTITION BY id ORDER BY id DESC
                ) AS [date]
            ,ROW_NUMBER() OVER (
                PARTITION BY DATE ORDER BY id
                ) seq
        FROM cte
        ) ddt
    ) Final
WHERE Final.RNO = 1
ORDER BY Final.RNO

它会起作用的

询问是否有任何疑问

;With cte( id, value ,date)
AS
(

 SELECT 1, 324,'2017-05-19 02:00:00' Union all
 SELECT 2, 564,'2017-05-19 04:00:00' Union all
 SELECT 3, 675,'2017-05-19 03:00:00' Union all
 SELECT 3, 642,'2017-05-19 05:00:00' Union all
 SELECT 4, 642,'2017-05-19 07:00:00' Union all
 SELECT 5, 642,'2017-05-19 06:00:00' Union all
 SELECT 6, 642,'2017-05-19 05:00:00' Union all
 SELECT 6, 642,'2017-05-19 12:00:00' Union all
 SELECT 7, 642,'2017-05-19 01:00:00' Union all
 SELECT 8, 642,'2017-05-19 13:00:00' Union all
 SELECT 8, 642,'2017-05-19 02:00:00' 
 )
SELECT id
    ,value
    ,DATE
FROM (
    SELECT *
        ,ROW_NUMBER() OVER (
            PARTITION BY id ORDER BY id
            ) RNO
    FROM (
        SELECT id
            ,value
            ,MAX([date]) OVER (
                PARTITION BY id ORDER BY id DESC
                ) AS [date]
            ,ROW_NUMBER() OVER (
                PARTITION BY DATE ORDER BY id
                ) seq
        FROM cte
        ) ddt
    ) Final
WHERE Final.RNO = 1
ORDER BY Final.RNO

输出

id  value   DATE
---------------------------
1   324     2017-05-19 02:00:00
2   564     2017-05-19 04:00:00
3   642     2017-05-19 05:00:00
4   642     2017-05-19 07:00:00
5   642     2017-05-19 06:00:00
6   642     2017-05-19 12:00:00
7   642     2017-05-19 01:00:00
8   642     2017-05-19 13:00:00

输出

id  value   DATE
---------------------------
1   324     2017-05-19 02:00:00
2   564     2017-05-19 04:00:00
3   642     2017-05-19 05:00:00
4   642     2017-05-19 07:00:00
5   642     2017-05-19 06:00:00
6   642     2017-05-19 12:00:00
7   642     2017-05-19 01:00:00
8   642     2017-05-19 13:00:00

结果是和startOk一样的表，我看不到不同的日期。正确的是
从name\u表组中按id、value选择id、value、max（date）
结果与startOk相同，我看不到不同的日期。正确的是，
从name\u table group by id中选择id、value、max（date），value
将返回id=3的两行，但只返回一行。将返回id=3的两行，但只返回一行。如果一个id有两个具有相同max time但“value”不同的RIW，则预期结果如何？您想要两行吗？DBMS:MSSQL，我需要一行用于一个idSQL服务器，然后您可以使用窗口函数！哪一行，如果一个id有日期限制？如果一个id有两个RIW，最大时间相同，但“值”不同，那么预期结果是什么？您想要两行吗？DBMS:MSSQL，我需要一行用于一个idSQL服务器，然后您可以使用窗口函数！哪一排，如果有身份证的约会领带？@Aleks很乐意帮忙：-）Aleks很乐意帮忙：-）Сббб102，бб15，бб1，бб30，ббббб15，我听不懂你的语言Сбб102，б15，我听不懂你的语言