从表中选择最新版本（MS SQL Server、SQLite）答案

【问题标题】：Select newest version from table (MS SQL Server, SQLite)从表中选择最新版本（MS SQL Server、SQLite）
【发布时间】：2015-12-04 10:47:43
【问题描述】：

比方说，我有下表：

ID | UID | Version | Content
---+-----+---------+-----------------------------
 1 |   1 |       1 | Something
 2 |   1 |       2 | Something (changed)
 3 |   2 |       1 | Alice has a cat
 4 |   2 |       2 | Alice has a cat and a dog
 5 |   2 |       3 | Alice has a cat and a canary

我需要创建查询，它将返回所有对象，但仅限于最新版本，所以在这种情况下：

ID | UID | Version | Content
---+-----+---------+-----------------------------
 2 |   1 |       2 | Something (changed)
 5 |   2 |       3 | Alice has a cat and a canary

由于 SQL 方言不同，我将在 MS Sql Server 2008 和 SQLite 3 上运行此查询。

我怎样才能做到这一点？

【问题讨论】：

标签： sql sql-server sqlite

【解决方案1】：

NOT EXISTS查询：

select *
from tablename t1
where not exists (select 1 from tablename t2
                  where t2.uid = t1.uid
                    and t2.version > t1.version)

JOIN查询：

select t1.*
from tablename t1
    join (select uid, max(version) as version from tablename group by uid) t2
    on t2.uid = t1.uid and t2.version = t1.version

相关子查询：

select t1.*
from tablename t1
where t1.version = (select max(version) from tablename t2
                    where t2.uid = t1.uid)

IN子查询：

select *
from tablename
where (uid, version) IN (select uid, max(version) from tablename
                         group by uid)

【讨论】：

【解决方案2】：

有几种不同的方法可以解决这个问题。但每种方法都遵循相同的原则。

查询的一部分将标识每个 UID 的最新版本。这将用于过滤记录集。

在下面的示例中，我使用子查询找到了每个 UID 的当前版本，然后我用它来过滤主记录集。

Example

/* This table gives us some sample records
 * to experiment with.
 */
DECLARE @Example TABLE
    (
        ID            INT                PRIMARY KEY,
        [UID]        INT                NOT NULL,
        [Version]    INT                NOT NULL,
        Content        VARCHAR(255)    NOT NULL
    )
;

/* Populate the sample data.
 */
INSERT INTO @Example
    (
        ID,
        [UID],
        [Version],
        Content
    )
VALUES
    (1, 1, 1, 'Something'),
    (2, 1, 2, 'Something (changed)'),
    (3, 2, 1, 'Alice has a cat'),
    (4, 2, 2, 'Alice has a cat and a dog'),
    (5, 2, 3, 'Alice has a cat and a canary')
;

/* Return the most recent version of each UID.
 * This is done by using a sub query to calcualte the max of each UIDs version.
 */
SELECT
    e.*
FROM
    @Example AS e
        INNER JOIN
            (    
                /* This sub query finds the max version of
                 * each UID.
                 */
                SELECT
                    [UID],
                    MAX([Version]) AS Current_Version
                FROM
                    @Example    
                GROUP BY
                    [UID]
            ) AS mv
        ON    mv.[UID]            = e.[UID]
        AND    mv.Current_Version    = e.[Version]
;

【讨论】：