BigQuery SQL：减少连续行计算中的剩余分钟数计数器[重复]

Question

我有一些带有日期和聚合 minutes 计数器时间的日志。

每个小时（一排）都被这样一个技巧填满：Duplicate groups of records to fill multiple date gaps in Google BigQuery

问题：

我想在 minutes 仍然可用的情况下完成 time 列（在此处了解，每小时最多 60 分钟）

这是所需的输出： remainingTime 是由前几行产生的。假设remainingTime = 70 分钟

#standardSQL
WITH history AS (
  SELECT '2017-01-01' AS date, 'a' AS product, 0 AS minutes UNION ALL
  SELECT '2017-01-02' AS date, 'a' AS product, 100 AS minutes UNION ALL
  SELECT '2017-01-03' AS date, 'a' AS product, 0 AS minutes UNION ALL
  SELECT '2017-01-04' AS date, 'a' AS product, 0 AS minutes UNION ALL
  SELECT '2017-01-05' AS date, 'a' AS product, 30 AS minutes UNION ALL
  SELECT '2017-01-06' AS date, 'a' AS product, 0 AS minutes UNION ALL
  SELECT '2017-01-01' AS date, 'b' AS product, 100 AS minutes UNION ALL
  SELECT '2017-01-02' AS date, 'b' AS product, 0 AS minutes UNION ALL
  SELECT '2017-01-03' AS date, 'b' AS product, 0 AS minutes UNION ALL
  SELECT '2017-01-04' AS date, 'b' AS product, 0 AS minutes UNION ALL
  SELECT '2017-01-05' AS date, 'b' AS product, 0 AS minutes 

+---------+------------+---------+---------------+---------------------+
| product |    date    | minutes | remainingTime |       time          |
+---------+------------+---------+---------------------------------+
|    a    | 2017-01-01 |   0     |  10           | 60 (max 60 reached) | // 0 + 70 - 60 = 10
|    a    | 2017-01-02 |   100   |  50           | 60 (same)           | // 100 + 10 - 60 = 50
|    a    | 2017-01-03 |   0     |  0            | 50 (only 50/60)     | // 0 + 50 - 50 = 0            
|    a    | 2017-01-04 |   0     |  0            | 0 (and so on)       | // 0 + 0 - 0 = 0
|    a    | 2017-01-05 |   30    |  0            | 30                  | // 30 + 0 - 30 = 0
|    a    | 2017-01-06 |   0     |  0            | 0                   | // 0 + 0 - 0 = 0
 

... and so on for other products
+---------------+--------+------+

我几乎完成了一个复杂而丑陋的查询，但我目前遇到了一个临时计算列..

（PS：我多年没有练习 SQL，所以我正在重新学习基础知识并同时发现 BigQuery 标准 SQL）

谢谢！

Answer 1

BigQuery 本身不支持递归操作。尝试将array_agg() 与JavaScript user-defined function 结合使用，但这种方法的可扩展性不是很高：

CREATE TEMP FUNCTION special_sum(x ARRAY<INT64>)
RETURNS INT64
LANGUAGE js
AS """
  var remaining = 70;
  var time = 0;
  for (const num of x)
  {
     time = Math.min(parseInt(num) + remaining, 60);
     remaining = parseInt(num) + remaining - time;
  }
  return time;
""";

WITH history AS (
  SELECT '2017-01-01' AS date, 'a' AS product, 0 AS minutes UNION ALL
  SELECT '2017-01-02' AS date, 'a' AS product, 100 AS minutes UNION ALL
  SELECT '2017-01-03' AS date, 'a' AS product, 0 AS minutes UNION ALL
  SELECT '2017-01-04' AS date, 'a' AS product, 0 AS minutes UNION ALL
  SELECT '2017-01-05' AS date, 'a' AS product, 30 AS minutes UNION ALL
  SELECT '2017-01-06' AS date, 'a' AS product, 0 AS minutes UNION ALL
  SELECT '2017-01-01' AS date, 'b' AS product, 100 AS minutes UNION ALL
  SELECT '2017-01-02' AS date, 'b' AS product, 0 AS minutes UNION ALL
  SELECT '2017-01-03' AS date, 'b' AS product, 0 AS minutes UNION ALL
  SELECT '2017-01-04' AS date, 'b' AS product, 0 AS minutes UNION ALL
  SELECT '2017-01-05' AS date, 'b' AS product, 0 AS minutes 
)
select *, 
 special_sum(array_agg(minutes) over (partition by product order by date rows unbounded preceding)) as time 
from history