找出条件 mysql 5.7 中每个用户的时差

Question

这是我的fiddlehttps://dbfiddle.uk/?rdbms=mysql_5.7&fiddle=7c549a3de0c8002ec43381462ba6a801

假设我有这样的数据

CREATE TABLE test (
  ID INT,
  user_id INT,
  createdAt DATE,
  status_id INT
);

INSERT INTO test VALUES
  (1, 12, '2020-01-01', 4),
  (2, 12, '2020-01-03', 7),
  (3, 12, '2020-01-06', 7),
  (4, 13, '2020-01-02', 5),
  (5, 13, '2020-01-03', 6),
  (6, 14, '2020-03-03', 8),
  (7, 13, '2020-03-04', 4),
  (8, 15, '2020-04-04', 7),
  (9, 14, '2020-03-02', 6),
  (10, 14, '2020-03-10', 5),
  (11, 13, '2020-04-10', 8);
  
select * from test
order by createdAt;

这是 table 之后 select (*)

+----+---------+------------+-----------+
| ID | user_id | createdAt  | status_id |
+----+---------+------------+-----------+
|  1 |      12 | 2020-01-01 |         4 |
|  4 |      13 | 2020-01-02 |         5 |
|  2 |      12 | 2020-01-03 |         7 |
|  5 |      13 | 2020-01-03 |         6 |
|  3 |      12 | 2020-01-06 |         7 |
|  9 |      14 | 2020-03-02 |         6 |
|  6 |      14 | 2020-03-03 |         8 |
|  7 |      13 | 2020-03-04 |         4 |
| 10 |      14 | 2020-03-10 |         5 |
|  8 |      15 | 2020-04-04 |         7 |
| 11 |      13 | 2020-04-10 |         8 |
+----+---------+------------+-----------+

id是交易的id，user_Id是进行交易的用户的id，createdAt是交易发生的日期，status_id是交易的状态（如果status_Id 为 7，则交易被拒绝或不批准。

所以在这种情况下，我想找出在“2020-02-01”到“2020-04-01”之间的时间范围内每个重复用户的每笔批准交易的时间差，重复用户是用户谁在时间范围结束前进行交易，并且至少在该时间范围内再次进行 1 次交易，在这种情况下，用户在 '2020-04-01' 之前进行批准交易并且至少再次进行 1 次批准交易在“2020-02-01”和“2020-04-01”之间。

根据解释，我使用了这个查询

SELECT SUM(transactions) AS transactions,
       MIN(`MIN`) AS `MIN`,
       MAX(`MAX`) AS `MAX`,
       SUM(total) / SUM(transactions) AS `AVG`
FROM (
  SELECT user_id,
         COUNT(*) AS transactions, 
         MIN(diff) AS `MIN`, 
         MAX(diff) AS `MAX`, 
         SUM(diff) AS total
  FROM (
    SELECT user_id, DATEDIFF((SELECT MIN(t2.createdAt)
                              FROM test t2
                              WHERE t2.user_id = t1.user_id
                                AND t1.createdAt < t2.createdAt
                                AND t2.status_id in (4, 5, 6, 8)
                              ), t1.createdAt) AS diff
    FROM test t1
    WHERE status_id in (4, 5, 6, 8)
    HAVING SUM(status_id != 7 and createdAt < '2020-04-01') > 1
               AND SUM(status_id != 7 AND createdAt BETWEEN '2020-02-01'
               AND '2020-04-01')
  ) DiffTable
  WHERE diff IS NOT NULL
  GROUP BY user_id
) totals

它说

In aggregated query without GROUP BY, expression #1 of SELECT list contains nonaggregated column 'db_314931870.t1.user_id'; this is incompatible with sql_mode=only_full_group_by

预期结果

+-----+-----+---------+
| MIN | MAX |   AVG   |
+-----+-----+---------+
|   1 |  61 | 21,6667 |
+-----+-----+---------+

解释：min（最小值）是 1 天的差异，发生在 users_id 14 人在“2020-03-02”进行审批交易并在“2020-03-03”再次进行审批交易,max(最大)是61时间差发生在users_Id13人在'2020-01-03'做审批交易并在'2020-03-04'再次进行审批交易，平均时间差是时间范围内所有时间差的总和：计算时间范围内发生的交易

Answer 1

SELECT MIN(DATEDIFF(t2.createdAt, t1.createdAt)) min_diff, 
       MAX(DATEDIFF(t2.createdAt, t1.createdAt)) max_diff, 
       AVG(DATEDIFF(t2.createdAt, t1.createdAt)) avg_diff
FROM test t1
JOIN test t2 ON t1.user_id = t2.user_id 
            AND t1.createdAt < t2.createdAt
            AND 7 NOT IN (t1.status_id, t2.status_id)
JOIN (SELECT t3.user_id
      FROM test t3
      WHERE t3.status_id != 7
      GROUP BY t3.user_id
      HAVING SUM(t3.createdAt < '2020-04-01')
         AND SUM(t3.createdAt BETWEEN '2020-02-01' AND '2020-04-01')) t4 ON t1.user_id = t4.user_id
WHERE NOT EXISTS (SELECT NULL
                  FROM test t5
                  WHERE t1.user_id = t5.user_id
                    AND t5.status_id != 7
                    AND t1.createdAt < t5.createdAt
                    AND t5.createdAt < t2.createdAt)

fiddle 有简短的解释。

找出条件 mysql 5.7 中每个用户的时差

find out time difference for every user in condition mysql 5.7

mysql

datetime

having

having-clause