如何计算当前值及其以下所有值的总和答案

【问题标题】：How to calculate sum for current value and all the values below it如何计算当前值及其以下所有值的总和
【发布时间】：2022-02-21 17:42:52
【问题描述】：

我有一个表格，其中包含有关特定版本的每个用户级别的信息。假设它的名称是 user_level_advanced。

version	user_id	level
0.9.3	1	2
0.9.5	2	3
0.9.3	3	4
0.9.3	4	5

并且我想计算每个级别的升级次数，以便每个用户都应考虑低于其当前级别的级别。对于上表，结果应该是这样的。

version	level	advanced_count
0.9.3	1	3
0.9.3	2	3
0.9.3	3	2
0.9.3	4	2
0.9.3	5	1
0.9.5	1	1
0.9.5	2	1
0.9.5	3	1

with user_level_advanced as(
    select 
        "0.9.3" as version, 1 as user_id, 2 as level_advanced_max
        union all
       select  "0.9.5" as version, 2 as user_id, 3 as level_advanced_max
        union all 
        select "0.9.3" as version, 3 as user_id, 4 as level_advanced_max
        union all 
        select "0.9.3" as version, 4 as user_id, 5 as level_advanced_max
),
user_grouped_by as
(
    select version, level_advanced_max, count(*) as level_advanced_count 
    from user_level_advanced 
    group by version, level_advanced_max
)

select  version,
        level_advanced_max,
        sum(level_advanced_count) over(partition by version order by level_advanced_max asc rows between current row and unbounded following)
from user_grouped_by

我使用这个查询来计算它，但它有一个缺陷，如果 user_level_advanced 表中缺少一个级别，那么结果表也会丢失它。感谢您的帮助。

version	level	advanced_count
0.9.3	2	3
0.9.3	4	2
0.9.3	5	1
0.9.5	3	1

【问题讨论】：

将结果留给列举了所有可能级别的表格。

标签： sql group-by google-bigquery

【解决方案1】：

考虑以下方法

select version, level, count(*) advanced_count
from user_level_advanced, 
unnest(generate_array(1, level_advanced_max)) level
group by version, level

如果应用于您问题中的样本数据 - 输出是

【讨论】：

【解决方案2】：

您可以使用cross join 生成所有可能的（级别、版本）对，然后将其加入您的表格以获得预期的结果：

with recursive levels as
(select 1 as level
union all
select level + 1 from levels
where level < (select max(level) from user_level_advanced)),
versions as
(select distinct version from user_level_advanced),
cte as
(select * from versions cross join levels)
select cte.version, cte.level, count(*)
from cte inner join user_level_advanced ula
on ula.level >= cte.level and ula.version = cte.version
group by cte.version, cte.level

Fiddle

【讨论】：