这是你需要的查询:

 SELECT b.id, a.home,b.[datetime],b.player,a.resource FROM
 (SELECT home,MAX(resource) AS resource FROM tbl_1 GROUP BY home) AS a

 LEFT JOIN

 (SELECT id,home,[datetime],player,resource FROM tbl_1) AS b
 ON  a.resource = b.resource WHERE a.home =b.home;

下面是T-SQL版本:

-- Test data
DECLARE @TestTable TABLE (id INT, home INT, date DATETIME, 
  player VARCHAR(20), resource INT)
INSERT INTO @TestTable
SELECT 1, 10, '2009-03-04', 'john', 399 UNION
SELECT 2, 11, '2009-03-04', 'juliet', 244 UNION
SELECT 5, 12, '2009-03-04', 'borat', 555 UNION
SELECT 3, 10, '2009-03-03', 'john', 300 UNION
SELECT 4, 11, '2009-03-03', 'juliet', 200 UNION
SELECT 6, 12, '2009-03-03', 'borat', 500 UNION
SELECT 7, 13, '2008-12-24', 'borat', 600 UNION
SELECT 8, 13, '2009-01-01', 'borat', 700

-- Answer
SELECT id, home, date, player, resource 
FROM (SELECT id, home, date, player, resource, 
    RANK() OVER (PARTITION BY home ORDER BY date DESC) N
    FROM @TestTable
)M WHERE N = 1

-- and if you really want only home with max date
SELECT T.id, T.home, T.date, T.player, T.resource 
    FROM @TestTable T
INNER JOIN 
(   SELECT TI.id, TI.home, TI.date, 
        RANK() OVER (PARTITION BY TI.home ORDER BY TI.date) N
    FROM @TestTable TI
    WHERE TI.date IN (SELECT MAX(TM.date) FROM @TestTable TM)
)TJ ON TJ.N = 1 AND T.id = TJ.id

编辑 不幸的是，MySQL中没有RANK() OVER函数。 但它可以被模拟，见模拟分析(AKA排名)函数与MySQL。 这是MySQL版本:

SELECT id, home, date, player, resource 
FROM TestTable AS t1 
WHERE 
    (SELECT COUNT(*) 
            FROM TestTable AS t2 
            WHERE t2.home = t1.home AND t2.date > t1.date
    ) = 0

另一种方法是gt每个组的最新行使用子查询，基本上计算每个组的每一行的排名，然后过滤出你的最新行，rank = 1

select a.*
from topten a
where (
  select count(*)
  from topten b
  where a.home = b.home
  and a.`datetime` < b.`datetime`
) +1 = 1

DEMO

为了更好地理解，这里是每行排名no的可视化演示

通过阅读一些评论，如果有两行具有相同的'home'和'datetime'字段值怎么办?

上述查询将失败，并将返回超过1行以上的情况。为了掩盖这种情况，需要另一个标准/参数/列来决定在上述情况下应该采取哪一行。通过查看样本数据集，我假设有一个主键列id，应该设置为自动递增。因此，我们可以使用这个列来选择最近的行，方法是在CASE语句的帮助下调整相同的查询

select a.*
from topten a
where (
  select count(*)
  from topten b
  where a.home = b.home
  and  case 
       when a.`datetime` = b.`datetime`
       then a.id < b.id
       else a.`datetime` < b.`datetime`
       end
) + 1 = 1

DEMO

上面的查询将在相同的datetime值中选择id最高的行

视觉演示的排名没有为每一行

(注意:Michael的答案非常适合目标列datetime不能为每个不同的home具有重复值的情况。)

如果你的表有重复的homexdatetime行，你只需要为每个不同的家列选择一行，这是我的解决方案:

您的表需要一个唯一的列(如id)。如果没有，创建一个视图并向其添加一个随机列。

使用此查询为每个唯一的主值选择单行。如果datetime重复，则选择最低的id。

SELECT tt.*
FROM topten tt
INNER JOIN
    (
    SELECT min(id) as min_id, home from topten tt2
    INNER JOIN 
        (
        SELECT home, MAX(datetime) AS MaxDateTime
        FROM topten
        GROUP BY home) groupedtt2
    ON tt2.home = groupedtt2.home
    ) as groupedtt
ON tt.id = groupedtt.id

试试这个

select * from mytable a join
(select home, max(datetime) datetime
from mytable
group by home) b
 on a.home = b.home and a.datetime = b.datetime

问候 K

试试这个SQL Server:

WITH cte AS (
   SELECT home, MAX(year) AS year FROM Table1 GROUP BY home
)
SELECT * FROM Table1 a INNER JOIN cte ON a.home = cte.home AND a.year = cte.year