如何选择最大值(column value)的行，与sql中的另一列不同？

xbp102n0 于 2021-06-25 发布在 Mysql

关注(0)|答案(19)|浏览(513)

我的table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 保持最大值 datetime .
结果是：

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。结果集有130行，但数据库有187行。结果包括的一些副本 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:

结果各不相同。

sql mysql greatest-n-per-group Max Distinct

来源：https://stackoverflow.com/questions/48244938/how-to-group-by-in-sql-with-latest-date-column

19条答案

按热度按时间

yws3nbqq16#

@michae这个被接受的答案在大多数情况下都能很好地工作，但在以下情况下，它失败了。
如果有两行具有相同的homeid和datetime，查询将返回这两行，而不是根据需要返回distinct homeid，对于add distinct in查询，如下所示。

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

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1tu0hz3e17#

这里是mysql版本，它只打印一个组中最大重复项（datetime）的条目。
你可以在这里测试http://www.sqlfiddle.com/#!2/0a4ae/1个

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)

接受答案的结果

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)

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gdrx4gfi18#

希望下面的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

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clj7thdc19#

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

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我来回答

如何选择最大值(column value)的行，与sql中的另一列不同？

19条答案

样本数据

带用户变量的mysql版本

接受答案的结果

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