posgresql中更宽的数据集-SQL

如何解决posgresql中更宽的数据集-SQL

我有下表：

CREATE TABLE my_table 
(
    the_debt_id varchar(6) NOT NULL,the_debt_paid date NOT NULL,the_debt_due date NOT NULL
)

INSERT INTO my_table
VALUES ('LMUS01','2019-05-03','2019-05-02'),('LMUS01','2019-06-03','2019-06-02'),'2019-07-01','2019-07-02'),('LMUS02','2019-05-07'),'2019-06-07','2019-06-07')

然后我提取了the_debt_paid的最后一条记录和倒数第二条记录

select * from
(
SELECT *,row_number()
OVER (PARTITION BY the_debt_id ORDER BY the_debt_paid DESC) as rn
FROM my_table
)A where rn<=2

所以我有这个结果：

the_debt_id    the_debt_paid     the_debt_due
LMUS01         2019-07-01        2019-07-02
LMUS01         2019-06-03        2019-06-02
LMUS02         2019-06-07        2019-06-07
LMUS02         2019-05-03        2019-05-07

是否有一种方法可以通过分支the_debt_paid来在SQL语句中获得更广泛的形式以获得预期的输出？

the_debt_id    the_debt_paid_last     the_debt_paid_next
LMUS01         2019-07-01             2019-06-03
LMUS02         2019-06-07             2019-05-03

我坚持使用以下代码：

select CASE 
WHEN rn = 1 THEN the_debt_paid_last = the_debt_paid 
WHEN rn = 2 THEN the_debt_paid_next = the_debt_paid
END
FROM (
select * FROM
(
SELECT *,row_number()
OVER (PARTITION BY the_debt_id ORDER BY the_debt_paid DESC) as rn
FROM my_table) A 
WHERE rn<=2) final

解决方法

您可以进行条件聚合：

select 
    the_debt_id,max(the_debt_paid) filter(where rn = 1) the_debt_paid_last,max(the_debt_paid) filter(where rn = 2) the_debt_paid_next
from (
    select 
        t.*,row_number() over (partition by the_debt_id order by the_debt_paid desc) as rn
    from my_table t
) a 
where rn <= 2
group by the_debt_id