如何计算每个用户mysql的timediff

Question

假设我有这样的数据表

ID  users_Id   createdAt
1   12         '2020-01-01'
2   12         '2020-01-03'
3   12         '2020-01-06'
4   13         '2020-01-02'
5   13         '2020-01-03'  

我如何获得每个交易和每个用户的时间差异，所以结果就像这样

MAX   MIN   AVERAGE    MEDIAN
3     1     3          3 

解释：

• 从 '2020-01-03' 到 '2020-01-06'（3 天）时，users_id 12 中出现的最大时间差异
• '2020-01-02' 和 '2020-01-03' 之间的交易时，users_id 13 中 timediff 的最小值发生
• 平均值为 3（users_Id 12 中 2 天 + users_Id 12 中 3 天 + users_Id 13 中 1 天）/users_id 计数（12 和 13）

Answer 1

你可以使用这样的东西（不计算中位数）：

SELECT MIN(diff) AS `MIN`, MAX(diff) AS `MAX`, SUM(diff) / COUNT(DISTINCT user_id) AS `AVG`
FROM (
  SELECT ID, user_id, DATEDIFF((SELECT t2.createdAt FROM test t2 WHERE t2.user_id = t1.user_id AND t1.createdAt <= t2.createdAt AND t2.id <> t1.id LIMIT 1), t1.createdAt) AS diff
  FROM test t1
  WHERE order_status_id in (4, 5, 6, 8)
) DiffTable
WHERE diff IS NOT NULL

---

在 MySQL 上计算中位数要复杂得多。但是您可以使用基于on this answer on StackOverflow 的类似内容。如您所见，查询变得非常混乱。 MySQL上没有像SUM或AVG这样的函数来获取中位数。

SELECT MIN(DiffTable.diff) AS `MIN`, MAX(DiffTable.diff) AS `MAX`, SUM(DiffTable.diff) / COUNT(DISTINCT user_id) AS `AVG`, MIN(median.diff) AS `MEDIAN`
FROM (
  SELECT ID, user_id, DATEDIFF((SELECT t2.createdAt FROM test t2 WHERE t2.user_id = t1.user_id AND t1.createdAt <= t2.createdAt AND t2.id <> t1.id LIMIT 1), t1.createdAt) AS diff
  FROM test t1
  WHERE order_status_id in (4, 5, 6, 8)
) DiffTable, (
  SELECT m1.diff FROM (
    SELECT ID, user_id, DATEDIFF((SELECT t2.createdAt FROM test t2 WHERE t2.user_id = t1.user_id AND t1.createdAt <= t2.createdAt AND t2.id <> t1.id LIMIT 1), t1.createdAt) AS diff
    FROM test t1
    WHERE order_status_id in (4, 5, 6, 8)
  ) m1, (
    SELECT ID, user_id, DATEDIFF((SELECT t2.createdAt FROM test t2 WHERE t2.user_id = t1.user_id AND t1.createdAt <= t2.createdAt AND t2.id <> t1.id LIMIT 1), t1.createdAt) AS diff
    FROM test t1
    WHERE order_status_id in (4, 5, 6, 8)
  ) m2
  WHERE m1.diff IS NOT NULL AND m2.diff IS NOT NULL
  GROUP BY m1.diff
  HAVING SUM(SIGN(1-SIGN(m1.diff-m2.diff))) = (COUNT(*)+1)/2
) median
WHERE DiffTable.diff IS NOT NULL

demo on dbfiddle.uk

Answer 2

在 MySQL created_at。这将为您提供除中位数之外的所有列：

select
    max(diff) max_diff,
    min(diff) min_diff,
    avg(diff) avg_diff
from (
    select
        t.*,
        datediff(
            created_at, 
            (select max(t1.created_at) from mytable t1 where t1.user_id = t.user_id and t1.created_at < t.created_at) 
        ) diff
    from mytable t
) t