sql - 如何在 PostgreSQL 中获得第二行？

Question

我有下一张 table :

COLUMN_NAME DATA_TYPE     PK    NULLABLE
inc_id      bigint        YES   NO
dt_cr       timestamp     NO    NO
email       varchar(255)  NO    YES

email 列不是唯一的，所以我应该使用 GROUP BY email。

问题: 如何从此表中获取最小日期、最小日期后的下一行以及与该日期相关的电子邮件？

我阅读了有关 first_value 和 nth_value 函数的信息，代码列在下面，但似乎无法正常工作。

SELECT J1.email, J2.first_date, J2.second_date 
FROM (
    SELECT email 
    FROM orders 
    GROUP BY email) J1
LEFT OUTER JOIN (
    SELECT email, 
        first_value(dt_cr) over (order by dt_cr ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) first_date, 
        NTH_VALUE(dt_cr, 2) over (order by dt_cr ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) second_date 
    FROM orders) J2  
ON (J1.email=J2.email);

有什么想法吗？

--已编辑 也许可以用 rank() 函数来完成..

Answer 1

此查询使用类似于子查询的 WITH 结构。在生产中使用之前使用 EXPLAIN 调查此查询，因为它在大表上可能会很慢:

WITH orders AS (
    SELECT 
          email
        , first_value(dt_cr) OVER wnd1 AS min_date
        , nth_value(dt_cr, 2) OVER wnd1 AS second_date 
    FROM orders
    WINDOW wnd1 AS ( PARTITION BY email ORDER BY email, dt_cr)
)
SELECT DISTINCT * 
FROM orders
WHERE second_date IS NOT NULL;