为由多列组成的每个组选择最大版本号

Question

我有以下表格：

应用程序

TYPE_ID | BUILD_ID | CONFIG_ID | VERSION_ID | (All foreign keys to the respective tables)
1       | 1        | 1         | 1          |
1       | 1        | 1         | 2          |
2       | 2        | 3         | 3          |
2       | 2        | 3         | 4          |

版本

ID | major | minor | patch
1  | 1     |0      |1
2  | 2     |0      |0
3  | 3     |0      |3
4  | 4     |0      |0

我需要从Apps 表中为TYPE_ID、BUILD_ID 和CONFIG_ID 的每个唯一组合选择最高版本的行。

版本号应按版本表中MAX(major * 1000000 + minor * 1000 + patch)计算。

因此，根据Apps 表的给定示例，结果将是：

TYPE_ID | BUILD_ID | CONFIG_ID | VERSION_ID |
1       | 1        | 1         | 2          |
2       | 2        | 3         | 4          |

尝试过这样的事情：

SELECT p1.* FROM Apps p1 
            INNER JOIN ( 
                SELECT max(VERSION_ID) MaxVersion, CONFIG_ID 
                FROM Apps  
                GROUP BY CONFIG_ID
            ) p2 
            ON p1.CONFIG_ID = p2.CONFIG_ID 
                AND p1.VERSION_ID = p2.MaxVersion  
            GROUP BY `TYPE_ID`, `BUILD_ID`, `CONFIG_ID`

但MAX 应用于VERSION_ID，我需要MAX 应用于major、minor 和patch 组合。

MySQL 版本 15.1 分发版 5.5.56-MariaDB

任何帮助将不胜感激。

干杯！

Answer 1

您可以使用问题中描述的公式计算每个type_id, build_id, config_id 的最大版本，再次使用相同的公式来定位版本：

SELECT sq.type_id, sq.build_id, sq.config_id, versions.id AS version_id_max
FROM (
    SELECT type_id, build_id, config_id, MAX(major * 1000000 + minor * 1000 + patch) AS max_version
    FROM apps
    INNER JOIN versions ON apps.version_id = versions.id
    GROUP BY type_id, build_id, config_id
) sq
INNER JOIN versions ON max_version = major * 1000000 + minor * 1000 + patch

+---------+----------+-----------+----------------+
| type_id | build_id | config_id | version_id_max |
+---------+----------+-----------+----------------+
|       1 |        1 |         1 |              2 |
|       2 |        2 |         3 |              4 |
+---------+----------+-----------+----------------+

Answer 2

试试这个：

select type_id, build_id, config_id,
      max(1000000*v.major+1000*v.minor+v.patch) as version 
from apps a left join versions v on a.version_id=v.id 
group by type_id, build_id, config_id

Answer 3

利用嵌套的派生子查询，以及标识VERSION_ID 的bit of hacky way 对应于MAX VERSION_NO。

我们基本上首先得到一个派生表，确定Apps 表中每一行的VERSION_NO。

现在使用该派生表作为SELECT 的源，我们按TYPE_ID、BUILD_ID 和CONFIG_ID 分组，并使用GROUP_CONCAT 和基于字符串操作的技巧，我们确定@987654332 @ 对应于最大 VERSION_NO，对于一个组。

尝试以下方法：

SELECT nest.TYPE_ID, 
       nest.BUILD_ID, 
       nest.CONFIG_ID, 
       SUBSTRING_INDEX(GROUP_CONCAT(DISTINCT nest.VERSION_ID   
                                    ORDER BY nest.VERSION_NO DESC  
                                    SEPARATOR ','), ',', 1) AS VERSION_ID 
FROM (
       SELECT A.TYPE_ID, 
              A.BUILD_ID, 
              A.CONFIG_ID, 
              A.VERSION_ID, 
              (V.major*1000000 + V.minor*1000 + V.patch) AS VERSION_NO 
       FROM Apps AS A
       INNER JOIN Versions AS V ON V.ID = A.VERSION_ID 
     ) AS nest 
GROUP BY nest.TYPE_ID, nest.BUILD_ID, nest.CONFIG_ID 

SQL FIDDLE

Answer 4

试试这个：

SELECT a1.type_id, a1.build_id, a1.config_id, a1.version_id
FROM apps a1
WHERE NOT EXISTS(
    (SELECT 'NEXT'
    FROM apps a2
    WHERE a2.type_id = a1.type_id
    AND a2.build_id = a1.build_id
    AND a2.config_id = a1.config_id
    AND a2.version_id > a1.version_id))

Answer 5

试试这个查询，我这里做什么，我模仿知名函数ROW_NUMBER() OVER (PARTITION BY Type_id, Build_id, Config_id ORDER BY major desc, minor desc, patch desc)。

select @type_id_lag := 0, @build_id_lag :=0, @config_id_lag := 0, @rn := 0;

select type_id, build_id, config_id, major, minor, patch from (
    select case when @type_id_lag = type_id and
                     @build_id_lag = build_id and
                     @config_id_lag = config_id then @rn := @rn + 1 else @rn := 1 rn,
           @type_id_lag := type_id type_id,
           @build_id_lag := build_id build_id,
           @config_id_lag := config_id config_id,
           v.major, v.minor, v.patch
    from Apps a
    left join Versions v on a.version_id = v.id
    order by a.type_id, a.build_id, a.config_id,
             v.major desc, v.minor desc, v.patch desc
) a where rn = 1;