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scala - 如何为每个 uniqueId 获取具有最大时间戳的行

转载 作者:行者123 更新时间:2023-12-02 03:15:38 25 4
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我有一个定义如下的表:

mysql> select * from rules;
+----+----------+-------+--------------+---------------+---------------------+
| id | uniqueId | name | json | definition | createdAt |
+----+----------+-------+--------------+---------------+---------------------+
| 1 | ABC | Rule1 | Json Repr v1 | Rule 1 Def v1 | 2016-05-13 09:10:44 |
| 2 | DEF | Rule2 | Json Repr v1 | Rule 2 Def v1 | 2016-05-13 09:12:04 |
| 3 | ABC | Rule1 | Json Repr v2 | Rule 1 Def v2 | 2016-05-13 09:14:04 |
| 4 | ABC | Rule1 | Json Repr v3 | Rule 1 Def v3 | 2016-05-13 09:14:15 |
| 5 | ABC | Rule1 | Json Repr v4 | Rule 1 Def v4 | 2016-05-13 09:14:22 |
| 6 | DEF | Rule2 | Json Repr v2 | Rule 2 Def v2 | 2016-05-13 09:14:31 |
+----+----------+-------+--------------+---------------+---------------------+

我知道我要运行的 SQL 查询来获取所需的行:

mysql> select * from rules where createdAt in (select max(createdAt) from rules group by uniqueId);
+----+----------+-------+--------------+---------------+---------------------+
| id | uniqueId | name | json | definition | createdAt |
+----+----------+-------+--------------+---------------+---------------------+
| 5 | ABC | Rule1 | Json Repr v4 | Rule 1 Def v4 | 2016-05-13 09:14:22 |
| 6 | DEF | Rule2 | Json Repr v2 | Rule 2 Def v2 | 2016-05-13 09:14:31 |
+----+----------+-------+--------------+---------------+---------------------+

我试图获得 Slick 查询,但我被卡住了。到目前为止,我可以按如下方式构建子查询:

val latestVersionOfRules = Rules.groupBy(p => p.uniqueId).map{ case(uniqueId, group) => (uniqueId, group.map(_.createdAt).max) }

但是,我无法访问 createdAt 值以与外部表连接。

我不确定如何构建上面的完整查询。

最佳答案

首先:您的查询原样并没有完全按照您的要求执行,因为它只是根据查询中的时间戳进行选择。如果您有多个具有最大时间戳的行,您将选择所有这些行。例如,如果您有:

mysql> select * from rules;
+----+----------+-------+--------------+---------------+---------------------+
| id | uniqueId | name | json | definition | createdAt |
+----+----------+-------+--------------+---------------+---------------------+
| 1 | ABC | Rule1 | Json Repr v1 | Rule 1 Def v1 | 2016-05-13 09:10:44 |
| 2 | DEF | Rule2 | Json Repr v1 | Rule 2 Def v1 | 2016-05-13 09:14:22 |
| 3 | ABC | Rule1 | Json Repr v2 | Rule 1 Def v2 | 2016-05-13 09:14:04 |
| 4 | ABC | Rule1 | Json Repr v3 | Rule 1 Def v3 | 2016-05-13 09:14:15 |
| 5 | ABC | Rule1 | Json Repr v4 | Rule 1 Def v4 | 2016-05-13 09:14:22 |
| 6 | DEF | Rule2 | Json Repr v2 | Rule 2 Def v2 | 2016-05-13 09:15:31 |
+----+----------+-------+--------------+---------------+---------------------+

请注意,第 2 行的时间戳等于 ABC 的最大时间戳。

你会回来:

mysql> select * from rules where createdAt in (select max(createdAt) from rules group by uniqueId);
+----+----------+-------+--------------+---------------+---------------------+
| id | uniqueId | name | json | definition | createdAt |
+----+----------+-------+--------------+---------------+---------------------+
| 2 | DEF | Rule2 | Json Repr v1 | Rule 2 Def v1 | 2016-05-13 09:14:22 |
| 5 | ABC | Rule1 | Json Repr v4 | Rule 1 Def v4 | 2016-05-13 09:14:22 |
| 6 | DEF | Rule2 | Json Repr v2 | Rule 2 Def v2 | 2016-05-13 09:15:31 |
+----+----------+-------+--------------+---------------+---------------------+

不是你想要的!在 SQL 中,您将加入一个子选择。这可以在 Slick 中完成:

// Create the query getting you the unique ID and max.
val latestVersionOfRules = Rules.groupBy(p => p.uniqueId).map {
case(uniqueId, group) => (uniqueId, group.map(_.createdAt).max)
}

// Join your table to the query. This is a DBIO of whatever your Rules
// result type is.
val results = Rules.join(latestVersionOfRules).on {
case (rulesRow, groupRow) =>
rulesRow.uniqueId === groupRow._1 &&
rulesRow.createdAt === groupRow._2
}.map {
case (rulesRow, _) => rulesRow
}.result

最后要注意的一件事:如果多行具有相同的最大时间戳,这将仍然为每个 uniqueId 返回多行。

关于scala - 如何为每个 uniqueId 获取具有最大时间戳的行,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37207886/

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