如何在kotlin暴露中选择别名列

Question

如何选择“按别名列分组”？我正在使用kotlin暴露。

fun getSubscribeInfoByRoleName(roleName: String): List<SubscribeInfo> {
        return SchemaSubscribersTable.join(SchemaVersionScheduleTable, JoinType.INNER, null, null) {
            SchemaSubscribersTable.schemaName eq SchemaVersionScheduleTable.schemaName }
                .slice(SchemaSubscribersTable.roleName, SchemaSubscribersTable.schemaName, SchemaVersionScheduleTable.version.max().alias("currentVersion"))
                .select { SchemaSubscribersTable.roleName.eq(roleName) and SchemaVersionScheduleTable.applyAt.less(CurrentDateTime()) }
                .groupBy(SchemaSubscribersTable.roleName,SchemaSubscribersTable.schemaName)
                .map {
                    SubscribeInfo(
                            roleName = it[SchemaSubscribersTable.roleName],
                            schemaName = it[SchemaSubscribersTable.schemaName],
                            currentVersion =it[/*How can I select currentVersion*/]
                    )
                }

    }

Answer 1

您应该将别名存储在变量中，然后使用它从ResultRow获取值：

val version = SchemaVersionScheduleTable.version.max().alias("currentVersion")
...
.map {
    SubscribeInfo(
        roleName = it[SchemaSubscribersTable.roleName],
        schemaName = it[SchemaSubscribersTable.schemaName],
        currentVersion = it[version]
    )
}