筛选有多少用户在 10 分钟内存款失败超过 5 次

Question

此查询返回每个用户在一小时内失败的存款次数。它仅过滤 5 个或更多失败的存款。 然而，现在我只需要过滤在 10 分钟内存款 5 笔或更多的用户。 我不知道该怎么做。

with windowed_data as
    (
    select *,
       max(window_count) over (partition by user_id, transaction_window) as max_window_count
from (select
          *,
          COUNT(*) OVER (PARTITION BY user_id, DATE_TRUNC('hour', created)  ORDER BY created) AS window_count,
          DATE_TRUNC('hour', created) AS transaction_window
      FROM source.payments2_transactions
      WHERE state = 'FAILED'
      and type = 'DEPOSIT'
      and country = 'PE'
      and created >= current_date - interval '120 days'
      and created < current_date + interval '1 day') wd)
select
    user_id,
    created,
    window_count,
    transaction_window,
    max_window_count
from windowed_data wd
where max_window_count >= 5
;

Answer 1

请尝试以下操作，我现在修改了查询，以识别在过去 120 天内的任何 10 分钟窗口内经历过 5 次或以上存款交易失败的用户，特别是在秘鲁的存款。它通过首先计算每笔交易的“

transaction_window

”来实现这一点，这将时间戳与最近的 10 分钟区块对齐。

WITH windowed_data AS (
    SELECT
        *,
        COUNT(*) OVER (PARTITION BY user_id, transaction_window) AS window_count,
        transaction_window
    FROM (
        SELECT
            user_id,
            created,
            DATE_TRUNC('minute', created) - (EXTRACT(MINUTE FROM created)::integer % 10) * INTERVAL '1 minute' AS transaction_window,
            state
        FROM
            source.payments2_transactions
        WHERE
            state = 'FAILED'
            AND type = 'DEPOSIT'
            AND country = 'PE'
            AND created >= CURRENT_DATE - INTERVAL '120 days'
            AND created < CURRENT_DATE + INTERVAL '1 day'
    ) AS subquery
),
aggregated_data AS (
    SELECT
        user_id,
        MAX(window_count) OVER (PARTITION BY user_id, transaction_window) AS max_window_count,
        transaction_window
    FROM
        windowed_data
)
SELECT
    user_id,
    transaction_window,
    max_window_count
FROM
    aggregated_data
WHERE
    max_window_count >= 5;