SQL (Snowflake) - 递归引用自身的列

Question

我看过很多关于这个主题的帖子，但找不到完全相同的帖子。

我想创建一个包含引用自身的列的视图。在下面的示例中：

值 1 + 2 是贯穿整个视图的静态值。
值 3 有更新的数据，所以我希望它在值 3 中有数据时执行值 3 +（值 1 - 值 2）。
当值3没有数据时，使用它的先验值计算。

Index	Value1	Value2	Value3	Calculation
1	0.5	0.25	1	1.25
2	0.5	0.25		1.5
3	0.5	0.25		1.75

这可能吗？到目前为止，我已经使用 LAG 或递归 CTE 计算索引 = 2 的行，但似乎无法弄清楚如何让它继续计算。

Answer 1

使用稍微扩展的示例数据来显示新的 value3 作为先例：

WITH data(Index, Value1, Value2, Value3) AS (
    SELECT * FROM VALUES   
    (1, 0.5, 0.25,  1),
    (2, 0.5, 0.25,  null),
    (3, 0.5, 0.25,  null),
    (4, 0.5, 0.25,  10),
    (5, 0.5, 0.25,  null),
    (6, 0.5, 0.25,  null),
    (7, 0.5, 0.25,  null)
)
select d.*
    ,iff(d.value3 is not null, d.index, null) as idx_k
    ,lag(d.value3) ignore nulls over(order by d.index) as l_value3
    ,lag(idx_k) ignore nulls over(order by d.index) as l_idx_k
    ,d.index - iff(idx_k is not null, idx_k, l_idx_k)+1 as r_d
    ,iff(d.value3 is not null, d.value3, l_value3) + r_d*(d.value1-d.value2) as calc
from data as d
order by 1;

给出：

INDEX	VALUE1	VALUE2	VALUE3	IDX_K	L_VALUE3	L_IDX_K	R_D	CALC
1	0.5	0.25	1	1			1	1.25
2	0.5	0.25			1	1	2	1.5
3	0.5	0.25			1	1	3	1.75
4	0.5	0.25	10	4	1	1	1	10.25
5	0.5	0.25			10	4	2	10.5
6	0.5	0.25			10	4	3	10.75
7	0.5	0.25			10	4	4	11

可以拼成：

select d.*
    ,iff(d.value3 is not null, d.value3, lag(d.value3) ignore nulls over(order by d.index)) + (d.index - iff(d.value3 is not null, iff(d.value3 is not null, d.index, null), lag(iff(d.value3 is not null, d.index, null)) ignore nulls over(order by d.index))+1)*(d.value1-d.value2) as calc
from data as d

如果你真的想要，但我更倾向于将其包装在 sub-select 中，只是为了让它呈现得更好：

select index, value1, value2, value3, calc 
from (
    select d.*
        ,iff(d.value3 is not null, d.index, null) as idx_k
        ,lag(d.value3) ignore nulls over(order by d.index) as l_value3
        ,lag(idx_k) ignore nulls over(order by d.index) as l_idx_k
        ,d.index - iff(idx_k is not null, idx_k, l_idx_k)+1 as r_d
        ,iff(d.value3 is not null, d.value3, l_value3) + r_d*(d.value1-d.value2) as calc
    from data as d
)
order by 1

递归 CTE：

WITH recursive r_cte as (
    select index, value1, value2, value3, value3 as calc
    from data 
    where index = 1
    
    union all
    
    select d.index, d.value1, d.value2, d.value3, iff(d.value3 is null, r.calc, d.value3) + d.value1 - d.value2 as calc
    from r_cte as r
    join data d on r.index + 1 = d.index
)
select * from r_cte

SQL (Snowflake) - 递归引用自身的列

SQL (Snowflake) - Column that recursively refers to itself

sql

snowflake-cloud-data-platform

递归 CTE：