下一个等级的 LAG() / LEAD() (Postgresql)答案

【问题标题】：LAG() / LEAD() of the next rank (Postgresql)下一个等级的 LAG() / LEAD() (Postgresql)
【发布时间】：2018-03-28 12:40:09
【问题描述】：

有什么方法可以强制 PostgreSQL LAG() 和 LEAD() 函数使用不是来自前行而是来自同一分区的下一个排名的值？

---------------------------------------------------
| client_id | order_id | product_id | year | rank |
---------------------------------------------------
|     1     |    1     |   111345   | 1995 |  1   |
|     1     |    1     |   912346   | 1995 |  1   |
|     1     |    1     |   212346   | 1995 |  1   |
|     1     |    2     |   233368   | 1998 |  4   |
|     1     |    2     |   133368   | 1998 |  4   |
|     1     |    3     |   412341   | 2005 |  6   |
|     2     |    55    |   312344   | 1995 |  1   |
|     2     |    57    |   812343   | 1999 |  2   |
---------------------------------------------------

预期结果是：

---------------------------------------------------------------------------
| client_id | order_id | product_id | year | rank | prev_year | next_year |
---------------------------------------------------------------------------
|     1     |    1     |   111345   | 1995 |  1   |    null   |   1998    |
|     1     |    1     |   912346   | 1995 |  1   |    null   |   1998    |
|     1     |    1     |   212346   | 1995 |  1   |    null   |   1998    |
|     1     |    2     |   233368   | 1998 |  4   |    1995   |   2005    |
|     1     |    2     |   133368   | 1998 |  4   |    1995   |   2005    |
|     1     |    3     |   412341   | 2005 |  6   |    1998   |   null    |
|     2     |    55    |   312344   | 1995 |  1   |    null   |   1999    |
|     2     |    57    |   812343   | 1999 |  2   |    1995   |   null    |    
---------------------------------------------------------------------------

如果year 在给定等级中有不同值，那么prev_year 和next_year 可以是这些值中的任何一个。例如：

---------------------------------------------------------------------------
| client_id | order_id | product_id | year | rank | prev_year | next_year |
---------------------------------------------------------------------------
|     1     |    1     |   111345   | 1994 |  1   |    null   |   1998    |
|     1     |    1     |   912346   | 1995 |  1   |    null   |   1998    |
|     1     |    1     |   212346   | 1996 |  1   |    null   |   1998    |
|     1     |    2     |   233368   | 1998 |  4   |    ????   |   null    |

???? 可以等于 1994、1995 或 1996

【问题讨论】：

标签： sql postgresql

【解决方案1】：

您应该在数据集上使用 lag() 和 lead() 函数，减少到每对 (client_id, rank) 的一行：

select 
    client_id, order_id, product_id, 
    t.year, rank, prev_year, next_year
from my_table t
join (
    select distinct on (client_id, rank)
        client_id, rank, year, 
        lag(year) over w as prev_year, 
        lead(year) over w as next_year
    from my_table
    window w as (partition by client_id order by rank)
    order by 1, 2, 3 desc
    ) s using (client_id, rank)
order by client_id, rank

 client_id | order_id | product_id | year | rank | prev_year | next_year 
-----------+----------+------------+------+------+-----------+-----------
         1 |        1 |     212346 | 1995 |    1 |           |      1998
         1 |        1 |     912346 | 1995 |    1 |           |      1998
         1 |        1 |     111345 | 1995 |    1 |           |      1998
         1 |        2 |     133368 | 1998 |    4 |      1995 |      2005
         1 |        2 |     233368 | 1998 |    4 |      1995 |      2005
         1 |        3 |     412341 | 2005 |    6 |      1998 |          
         2 |       55 |     312344 | 1995 |    1 |           |      1999
         2 |       57 |     812343 | 1999 |    2 |      1995 |          
(8 rows)

【讨论】：