如何在带有 where 子句的 sql 查询中获取最新的日期时间记录？

Question

我有两个表：RO_LAMEL_DATA 和 RO_MAIN_TABLE。 RO_MAIN_TABLE 包含具有记录密钥 (record_key) 的作品的所有序列号 (serial_nr)。 RO_LAMEL_DATA 每个记录键都有几条记录（在同一天），例如机器状态 (machine_status) 和日期时间值 (pr_date_time)。我想获得一个产品的最新机器状态。为此，我这样做：

select a.machine_status
from ro_lamel_Data a inner join (
    select record_key, max(pr_date_time) as MaxDate
    from ro_lamel_Data
    group by record_key
) ro_main_table on (a.record_key = ro_main_table.record_key) and a.pr_date_time = MaxDate
where a.record_key =(
    select record_key from ro_main_table where serial_nr = 'Y39489');

但是我得到错误：

single-row subquery returns more than one row

我该如何解决这个问题？提前致谢！

Answer 1

也许你需要这样的东西

WITH cte AS ( SELECT machine_status, 
                     record_key,
                     ROW_NUMBER() OVER (PARTITION BY record_key 
                                        ORDER BY pr_date_time DESC) rn
              FROM ro_lamel_Data )
SELECT cte.record_key, cte.machine_status last_status
FROM cte 
JOIN ro_main_table ON cte.record_key = ro_main_table.record_key
WHERE rn = 1

fiddle

Answer 2

如果你想要一行，使用 order by 和 fetch first:

select ld.machine_status
from ro_lamel_Data ld join
     ro_main_table mt
     using (record_key)
where mt.serial_nr = 'Y39489'
order by ld.pr_ate_time desc
fetch first 1 row only;

如何在带有 where 子句的 sql 查询中获取最新的日期时间记录？

How to get the latest date time record in an sql query with a where clause?

sql

oracle

greatest-n-per-group