Question

假设我有一个返回生产作业表的查询，在一列中，我在过去7天内每个作业都有一个输出数组：

 sku | job | outputs
-----------------------------
 A1  | 123 | {2,4,6,5,5,5,5}
 A1  | 135 | {0,0,0,3,5,7,9}
 B3  | 109 | {3,2,3,2,3,2,3}
 C5  | 144 | {5,5,5,5,5,5,5}

如何编写按SKU分组的查询（产品编号）并按位置汇总7天的输出？在这种情况下，您可以看到产品A1有两个生产作业：这些作业应合并到结果的一行中：

 sku | outputs
--------------------------
 A1  | {2,4,6,8,10,12,14}
 B3  | {3,2,3,2,3,2,3}
 C5  | {5,5,5,5,5,5,5}

Answer 1

你应该删除带有普通性的数组，按sku和常规计算组中元素的总和，最后使用sku中的组中的常数将总和聚合成数组：

select sku, array_agg(elem order by ordinality) as outputs
from (
    select sku, ordinality, sum(elem) as elem
    from jobs
    cross join unnest(outputs) with ordinality as u(elem, ordinality)
    group by 1, 2
    ) s
group by 1
order by 1

DbFiddle.

如果您经常在各种环境中使用此功能，则创建自定义聚合可能是合理的：

create or replace function sum_int_arrays(int[], int[])
returns int[] language sql immutable as $$
    select array_agg(coalesce(a, 0)+ b)
    from unnest($1, $2) as u(a, b)
$$;

create aggregate sum_int_array_agg(integer[]) (
    sfunc = sum_int_arrays,
    stype = int[]
);

select sku, sum_int_array_agg(outputs)
from jobs
group by 1
order by 1

DbFiddle.

如何按位置对postrgresql数组的元素求和？

1 个答案: