Question

SELECT * FROM ABC_CUSTOMER_DETAILS abc_detail
INNER JOIN ABC_CUSTOMERS abc_cust
ON abc_detail.ID=abc_cust.CUSTOMER_ID
WHERE abc_detail.COUNTRY_CODE='KE'
AND CREATION_TIMESTAMP=(SELECT MIN (CREATION_TIMESTAMP)
                        FROM ABC_CUSTOMER_DETAILS abc_detail
                        INNER JOIN ABC_CUSTOMERS abc_cust
                        ON abc_detail.ID=abc_cust.CUSTOMER_ID
                        WHERE abc_detail.COUNTRY_CODE='KE');

以上脚本查询从ABC_CUSTOMER_DETAILS到ABC_CUSTOMERS的联接记录，并选择具有最早时间戳的记录。

无论如何，如果我不能在JOIN条件下重复相同的WHERE和CREATION_TIMESTAMP条款？

Answer 1

有几种方法可以获得最早的记录，并避免两次输入相同的标准。

使用FETCH FIRST ROWS（从Oracle 12c开始提供）

select * 
from abc_customer_details cd
join abc_customers c on c.id = cd.customer_id
where cd.country_code = 'KE'
order by creation_timestamp
fetch first row only;

使用CTE（WITH子句）

with cte as
(
  select * 
  from abc_customer_details cd
  join abc_customers c on c.id = cd.customer_id
  where cd.country_code = 'KE'
)
select *
from cte
where (creation_timestamp) = (select min(creation_timestamp) from cte);

使用窗口功能

select *
from
(
  select cd.*, c.*, min(creation_timestamp) over () as min_creation_timestamp
  from abc_customer_details cd
  join abc_customers c on c.id = cd.customer_id
  where cd.country_code = 'KE'
)
where creation_timestamp = min_creation_timestamp;

（顺便说一下，我更改了所有这些查询中的加入条件。您似乎不太可能想加入abc_customer_details.id = abc_customers.customer_id。）

Answer 2

您可以使用MIN()分析函数。

SELECT
    *
FROM
    (
        SELECT
            abc_detail.*,
            abc_cust.*,
            MIN(creation_timestamp) OVER(
                PARTITION BY abc_detail.id
            ) AS min_timestamp
        FROM
            abc_customer_details abc_detail
            INNER JOIN abc_customers abc_cust
        ON abc_detail.id = abc_cust.customer_id
        WHERE
            abc_detail.country_code = 'KE'
    )
WHERE
    creation_timestamp = min_timestamp;

使用聚合函数根据MIN时间戳过滤记录

2 个答案: