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mysql - 如何计算每个用户mysql的timediff

问题描述

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'

我如何获得每笔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：

标签： mysqlsqldateselect

解决方案

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。但是您可以根据StackOverflow 上的这个答案使用类似的东西。如您所见，查询变得非常混乱。没有类似SUM或AVG在 MySQL 上获取中位数的函数。

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

dbfiddle.uk 上的演示

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