BigQuery：滾動每日計數訪問者的付款摘要

Question

我有這個數據：

日期	visitor_id	總付款
2022-01-01	一種	20
2022-01-01	乙	15
2022-01-01	C	20
2022-01-02	乙	10
2022-01-02	丁	25

我希望每天有 total_payment 等於或大於 20 美元的訪問者數量，話雖如此，我希望的結果是：

日期	計數訪問者
2022-01-01	2個
2022-01-02	4個

2022-01-01 是 2，因為只有 A 和 C 的付款超過 20 美元，但是在 2022-01-02 又增加了 2，因為 B 是 35 美元（總和），D 是 25 美元。

對此有任何可能的查詢嗎？ 我希望我的描述很清楚。 先感謝您。

Answer 1

您可以使用此查詢作為解決方案。 首先，我計算每個用戶的累計付款。 然后，我找到每個用戶超過 20$ 累計付款的最短日期。 在最后一步，我計算每個最短日期的用戶數，並累加該數字。 在 output 中，您不必有 first_day_users 列，但我保留它是為了更容易理解代碼。

所以 output 看起來像這樣：

WITH 
data AS(
  SELECT "2022-01-01" AS date, "A" AS visitor_id, 20 AS total_payment UNION ALL
  SELECT "2022-01-01" AS date, "B" AS visitor_id, 15 AS total_payment UNION ALL
  SELECT "2022-01-01" AS date, "C" AS visitor_id, 20 AS total_payment UNION ALL
  SELECT "2022-01-02" AS date, "B" AS visitor_id, 10 AS total_payment UNION ALL
  SELECT "2022-01-02" AS date, "D" AS visitor_id, 25 AS total_payment
),
user_cumulatives as 
(
    SELECT
        visitor_id,
        date,
        SUM(total_payment) OVER (PARTITION BY visitor_id ORDER BY date) as cumulative_payment
    FROM data
),
user_first_dates as 
(
    select visitor_id, min(date) as date
    from user_cumulatives
    where cumulative_payment >= 20
    group by 1
)
select date, count(*) as first_day_users, sum(count(*)) over (order by date) as count_visitor
from user_first_dates
group by 1
order by date

Answer 2

歡迎@Indri

下面的查詢將為您提供每天的行總和，其中total_amount大於等於 20，我相信這應該會給您您正在尋找的答案：

WITH data AS(
  SELECT "2022-01-01" AS date, "A" AS visitor_id, 20 AS total_payment
  UNION ALL
  SELECT "2022-01-01" AS date, "B" AS visitor_id, 15 AS total_payment
  UNION ALL
  SELECT "2022-01-01" AS date, "C" AS visitor_id, 20 AS total_payment
  UNION ALL
  SELECT "2022-01-02" AS date, "A" AS visitor_id, 10 AS total_payment
  UNION ALL
  SELECT "2022-01-02" AS date, "D" AS visitor_id, 25 AS total_payment
)

SELECT
*,
COUNT(*) OVER(ORDER BY date)
FROM data
WHERE total_payment >= 20