我的表是：
id home datetime player resource
---|-----|------------|--------|---------
1 | 10 | 04/03/2009 | john | 399
2 | 11 | 04/03/2009 | juliet | 244
5 | 12 | 04/03/2009 | borat | 555
3 | 10 | 03/03/2009 | john | 300
4 | 11 | 03/03/2009 | juliet | 200
6 | 12 | 03/03/2009 | borat | 500
7 | 13 | 24/12/2008 | borat | 600
8 | 13 | 01/01/2009 | borat | 700
我需要选择保持home最大值的每个不同datetime。
结果将是：
id home datetime player resource
---|-----|------------|--------|---------
1 | 10 | 04/03/2009 | john | 399
2 | 11 | 04/03/2009 | juliet | 244
5 | 12 | 04/03/2009 | borat | 555
8 | 13 | 01/01/2009 | borat | 700
我尝试过：
-- 1 ..by the MySQL manual:
SELECT DISTINCT
home,
id,
datetime AS dt,
player,
resource
FROM topten t1
WHERE datetime = (SELECT
MAX(t2.datetime)
FROM topten t2
GROUP BY home)
GROUP BY datetime
ORDER BY datetime DESC
不起作用。尽管数据库保持187，但结果集有130行。
结果包含一些重复的home。
-- 2 ..join
SELECT
s1.id,
s1.home,
s1.datetime,
s1.player,
s1.resource
FROM topten s1
JOIN (SELECT
id,
MAX(datetime) AS dt
FROM topten
GROUP BY id) AS s2
ON s1.id = s2.id
ORDER BY datetime
不。提供所有记录。
-- 3 ..something exotic:
有各种结果。
答案 0 :(得分：860)
你真是太近了！您需要做的就是选择主页及其最长日期时间，然后返回到两个字段上的topten表：
SELECT tt.*
FROM topten tt
INNER JOIN
(SELECT home, MAX(datetime) AS MaxDateTime
FROM topten
GROUP BY home) groupedtt
ON tt.home = groupedtt.home
AND tt.datetime = groupedtt.MaxDateTime
答案 1 :(得分：70)
这里有 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
修改
不幸的是，MySQL中没有RANK（）OVER函数
但可以模仿，见Emulating Analytic (AKA Ranking) Functions with MySQL
所以这是 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
答案 2 :(得分：64)
最快的MySQL解决方案，没有内部查询且没有GROUP BY：
SELECT m.* -- get the row that contains the max value
FROM topten m -- "m" from "max"
LEFT JOIN topten b -- "b" from "bigger"
ON m.home = b.home -- match "max" row with "bigger" row by `home`
AND m.datetime < b.datetime -- want "bigger" than "max"
WHERE b.datetime IS NULL -- keep only if there is no bigger than max
<强>解释：
使用home列自行加入表格。使用LEFT JOIN可确保表m中的所有行都显示在结果集中。表b中没有匹配项的NULL列{}为b。
JOIN上的其他条件要求仅匹配b列中datetime列上的值大于m中的行的LEFT JOIN行。
使用问题中发布的数据，+------------------------------------------+--------------------------------+
| the row from `m` | the matching row from `b` |
|------------------------------------------|--------------------------------|
| id home datetime player resource | id home datetime ... |
|----|-----|------------|--------|---------|------|------|------------|-----|
| 1 | 10 | 04/03/2009 | john | 399 | NULL | NULL | NULL | ... | *
| 2 | 11 | 04/03/2009 | juliet | 244 | NULL | NULL | NULL | ... | *
| 5 | 12 | 04/03/2009 | borat | 555 | NULL | NULL | NULL | ... | *
| 3 | 10 | 03/03/2009 | john | 300 | 1 | 10 | 04/03/2009 | ... |
| 4 | 11 | 03/03/2009 | juliet | 200 | 2 | 11 | 04/03/2009 | ... |
| 6 | 12 | 03/03/2009 | borat | 500 | 5 | 12 | 04/03/2009 | ... |
| 7 | 13 | 24/12/2008 | borat | 600 | 8 | 13 | 01/01/2009 | ... |
| 8 | 13 | 01/01/2009 | borat | 700 | NULL | NULL | NULL | ... | *
+------------------------------------------+--------------------------------+
将生成此对：
WHERE最后，NULL子句只保留b列中*的对（在上表中标有JOIN）;这意味着，由于m子句中的第二个条件，从datetime中选择的行在{{1}}列中具有最大值。
阅读SQL Antipatterns: Avoiding the Pitfalls of Database Programming书籍了解其他SQL技巧。
答案 3 :(得分：26)
即使每个home有两行或多行，并且DATETIME位相同，这也会有效：
SELECT id, home, datetime, player, resource
FROM (
SELECT (
SELECT id
FROM topten ti
WHERE ti.home = t1.home
ORDER BY
ti.datetime DESC
LIMIT 1
) lid
FROM (
SELECT DISTINCT home
FROM topten
) t1
) ro, topten t2
WHERE t2.id = ro.lid
答案 4 :(得分：23)
我认为这会给你想要的结果：
SELECT home, MAX(datetime)
FROM my_table
GROUP BY home
但如果您还需要其他列，只需与原始表进行联接（查看Michael La Voie answer）
最好的问候。
答案 5 :(得分：16)
由于人们似乎继续遇到这个帖子（评论日期范围从1。5年起）并不是那么简单：
SELECT * FROM (SELECT * FROM topten ORDER BY datetime DESC) tmp GROUP BY home
不需要聚合功能......
干杯。
答案 6 :(得分：10)
您也可以尝试这一个，对于大型表，查询性能会更好。它适用于每个家庭的记录不超过两个，并且它们的日期不同。更好的一般MySQL查询是上面的Michael La Voie的一个。
SELECT t1.id, t1.home, t1.date, t1.player, t1.resource
FROM t_scores_1 t1
INNER JOIN t_scores_1 t2
ON t1.home = t2.home
WHERE t1.date > t2.date
或者在Postgres或提供分析功能的dbs的情况下尝试
SELECT t.* FROM
(SELECT t1.id, t1.home, t1.date, t1.player, t1.resource
, row_number() over (partition by t1.home order by t1.date desc) rw
FROM topten t1
INNER JOIN topten t2
ON t1.home = t2.home
WHERE t1.date > t2.date
) t
WHERE t.rw = 1
答案 7 :(得分：8)
这适用于Oracle：
with table_max as(
select id
, home
, datetime
, player
, resource
, max(home) over (partition by home) maxhome
from table
)
select id
, home
, datetime
, player
, resource
from table_max
where home = maxhome
答案 8 :(得分：7)
尝试使用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
答案 9 :(得分：7)
SELECT tt.*
FROM TestTable tt
INNER JOIN
(
SELECT coord, MAX(datetime) AS MaxDateTime
FROM rapsa
GROUP BY
krd
) groupedtt
ON tt.coord = groupedtt.coord
AND tt.datetime = groupedtt.MaxDateTime
答案 10 :(得分：4)
SELECT c1, c2, c3, c4, c5 FROM table1 WHERE c3 = (select max(c3) from table)
SELECT * FROM table1 WHERE c3 = (select max(c3) from table1)
答案 11 :(得分：4)
这是MySQL版本，它只打印一个条目中存在重复MAX（日期时间）的条目。
你可以在这里测试http://www.sqlfiddle.com/#!2/0a4ae/1
mysql> SELECT * from topten;
+------+------+---------------------+--------+----------+
| id | home | datetime | player | resource |
+------+------+---------------------+--------+----------+
| 1 | 10 | 2009-04-03 00:00:00 | john | 399 |
| 2 | 11 | 2009-04-03 00:00:00 | juliet | 244 |
| 3 | 10 | 2009-03-03 00:00:00 | john | 300 |
| 4 | 11 | 2009-03-03 00:00:00 | juliet | 200 |
| 5 | 12 | 2009-04-03 00:00:00 | borat | 555 |
| 6 | 12 | 2009-03-03 00:00:00 | borat | 500 |
| 7 | 13 | 2008-12-24 00:00:00 | borat | 600 |
| 8 | 13 | 2009-01-01 00:00:00 | borat | 700 |
| 9 | 10 | 2009-04-03 00:00:00 | borat | 700 |
| 10 | 11 | 2009-04-03 00:00:00 | borat | 700 |
| 12 | 12 | 2009-04-03 00:00:00 | borat | 700 |
+------+------+---------------------+--------+----------+
SELECT *
FROM (
SELECT ord.*,
IF (@prev_home = ord.home, 0, 1) AS is_first_appear,
@prev_home := ord.home
FROM (
SELECT t1.id, t1.home, t1.player, t1.resource
FROM topten t1
INNER JOIN (
SELECT home, MAX(datetime) AS mx_dt
FROM topten
GROUP BY home
) x ON t1.home = x.home AND t1.datetime = x.mx_dt
ORDER BY home
) ord, (SELECT @prev_home := 0, @seq := 0) init
) y
WHERE is_first_appear = 1;
+------+------+--------+----------+-----------------+------------------------+
| id | home | player | resource | is_first_appear | @prev_home := ord.home |
+------+------+--------+----------+-----------------+------------------------+
| 9 | 10 | borat | 700 | 1 | 10 |
| 10 | 11 | borat | 700 | 1 | 11 |
| 12 | 12 | borat | 700 | 1 | 12 |
| 8 | 13 | borat | 700 | 1 | 13 |
+------+------+--------+----------+-----------------+------------------------+
4 rows in set (0.00 sec)
SELECT tt.*
FROM topten tt
INNER JOIN
(
SELECT home, MAX(datetime) AS MaxDateTime
FROM topten
GROUP BY home
) groupedtt ON tt.home = groupedtt.home AND tt.datetime = groupedtt.MaxDateTime
+------+------+---------------------+--------+----------+
| id | home | datetime | player | resource |
+------+------+---------------------+--------+----------+
| 1 | 10 | 2009-04-03 00:00:00 | john | 399 |
| 2 | 11 | 2009-04-03 00:00:00 | juliet | 244 |
| 5 | 12 | 2009-04-03 00:00:00 | borat | 555 |
| 8 | 13 | 2009-01-01 00:00:00 | borat | 700 |
| 9 | 10 | 2009-04-03 00:00:00 | borat | 700 |
| 10 | 11 | 2009-04-03 00:00:00 | borat | 700 |
| 12 | 12 | 2009-04-03 00:00:00 | borat | 700 |
+------+------+---------------------+--------+----------+
7 rows in set (0.00 sec)
答案 12 :(得分：4)
使用子查询对每组最近一行进行gt的另一种方法，该查询基本上计算每组每行的排名，然后按照rank = 1筛选出最近的行
select a.*
from topten a
where (
select count(*)
from topten b
where a.home = b.home
and a.`datetime` < b.`datetime`
) +1 = 1
为了更好地理解，以下是每行排名号的visual demo
通过阅读一些评论如果有两行具有相同的＆＃39; home＆＃39;和＆＃39; datetime＆＃39;字段值？
以上查询将失败，并将针对上述情况返回超过1行。为了掩盖这种情况，将需要另一个标准/参数/列来决定应该采取哪一行属于上述情况。通过查看样本数据集，我假设有一个主键列id应该设置为自动增量。因此，我们可以使用此列通过在CASE语句的帮助下调整相同的查询来选择最近的行，如
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
以上查询将选择具有相同datetime值
visual demo，每行的排名为
答案 13 :(得分：1)
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
此致 ķ
答案 14 :(得分：1)
为什么不使用： SELECT home，MAX（datetime）AS MaxDateTime，player，resource FROM topten GROUP BY home 我错过了什么吗？
答案 15 :(得分：1)
希望以下查询将给出所需的输出：
Select id, home,datetime,player,resource, row_number() over (Partition by home ORDER by datetime desc) as rownum from tablename where rownum=1
答案 16 :(得分：0)
这是您需要的查询：
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;
答案 17 :(得分：0)
@Michae在大多数情况下，已接受的答案将正常工作，但如果出现以下情况则会失败。
如果有2行具有HomeID和Datetime相同，则查询将返回两行，而不是根据需要返回不同的HomeID，如下所示在查询中添加Distinct。
SELECT DISTINCT tt.home , tt.MaxDateTime
FROM topten tt
INNER JOIN
(SELECT home, MAX(datetime) AS MaxDateTime
FROM topten
GROUP BY home) groupedtt
ON tt.home = groupedtt.home
AND tt.datetime = groupedtt.MaxDateTime
答案 18 :(得分：0)
（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。）
如果您的表中有{strong> home x datetime的重复行，并且您只需要为每个home单独的列选择一行，这就是我的解决方案对此：
您的表需要一个唯一的列（如id）。如果没有，请创建一个视图并向其中添加随机列。
使用此查询为每个唯一的home值选择一行。如果重复id，则选择最低的datetime。
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
答案 19 :(得分：0)
在 MySQL 8.0 中，这可以通过使用带有公共表表达式的 row_number() 窗口函数来有效地实现。
（这里的 row_number() 基本上为每个从 1 开始按资源降序排列的玩家的每一行生成唯一的序列。因此，对于序列号为 1 的每个玩家行将具有最高的资源值。现在我们需要做的就是正在为每个玩家选择序列号为 1 的行。可以通过围绕此查询编写外部查询来完成。但我们改用公共表表达式，因为它更具可读性。）
架构：
create TABLE TestTable(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
查询：
with cte as
(
select id, home, date , player, resource,
Row_Number()Over(Partition by home order by date desc) rownumber from TestTable
)
select id, home, date , player, resource from cte where rownumber=1
输出：
| id | home | 日期 | 玩家 | 资源 |
|---|---|---|---|---|
| 1 | 10 | 2009-03-04 00:00:00 | 约翰 | 399 |
| 2 | 11 | 2009-03-04 00:00:00 | 朱丽叶 | 244 |
| 5 | 12 | 2009-03-04 00:00:00 | borat | 555 |
| 8 | 13 | 2009-01-01 00:00:00 | borat | 700 |
db<>fiddle here
答案 20 :(得分：0)
如果有 2 条日期和家相同的记录，则接受的答案对我不起作用。加入后会返回2条记录。虽然我需要选择其中的任何（随机）。此查询用作连接的子查询，因此在那里不可能只限制 1。 这是我达到预期结果的方式。但是不知道性能。
select SUBSTRING_INDEX(GROUP_CONCAT(id order by datetime desc separator ','),',',1) as id, home, MAX(datetime) as 'datetime'
from topten
group by (home)