如何通过SQL中的另一列选择MAX(列值),DISTINCT的行?
-
03-07-2019 - |
题
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
我需要选择每个不同的 home
,其中包含 datetime
的最大值。
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。
解决方案
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
其他提示
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
修改强>结果
不幸的是,MySQL中没有RANK()OVER函数。
但它可以模拟,参见使用MySQL模拟分析(AKA排名)函数。
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
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>解释强>:
使用家代码>列加入表本身。使用
LEFT JOIN
可确保表 m
中的所有行都出现在结果集中。表 b
中没有匹配项的那些 NULL
用于 b
的列。
JOIN
上的另一个条件要求仅匹配 b
列中 b>
列中的行,而不是来自 datetime
列的行<代码>米代码>
使用问题中发布的数据, LEFT JOIN
将生成此对:
+------------------------------------------+--------------------------------+
| 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 | ... | *
+------------------------------------------+--------------------------------+
最后, WHERE
子句只保留 b
列中 NULL
的对(它们用标记) *
在上表中);这意味着,由于 JOIN
子句中的第二个条件,从 m
中选择的行在 datetime
列中具有最大值。
阅读 SQL Antipatterns:避免数据库编程的陷阱一书,了解其他SQL技巧
即使每个 home
有两行或更多行,并且等于 DATETIME
,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
我认为这会给你想要的结果:
SELECT home, MAX(datetime)
FROM my_table
GROUP BY home
但是如果您还需要其他列,只需与原始表进行连接(查看 Michael La Voie
答案)
最好的问候。
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。
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
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
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
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
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)
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
样本数据
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 |
+------+------+---------------------+--------+----------+
带有用户变量
的MySQL版本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)
接受答案'outout
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)
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
以下是视觉演示,每行排名为“否”,以便更好地理解
通过阅读一些评论如果有两行具有相同的'home'和'datetime'字段值会怎么样?
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
,应设置为自动递增。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
上面的查询将在相同的 datetime
值
视觉演示,每行排名为
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
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?
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;
@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