如何使用 mysql 找出每个用户每笔交易的时差（当天）

Question

我有一个 table 这样的：

CREATE TABLE test (
  ID SERIAL PRIMARY KEY,
  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);

this is my fiddle

其中 table 每笔交易的 ID 有 id，user_id 是用户，createdAt 是交易发生的日期，status_id 是每笔交易的状态（在这种情况下 status_id 4、5、6、8 是已批准的交易）

我想找出在“2020-02-01”和“2020-04-01”之间进行交易并批准了超过 1 笔交易的每个用户的每笔交易的最大、最小、平均不同日期时期

这是我的查询：

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 
  status_id in (4, 5, 6, 8)
  HAVING SUM(t1.user_id BETWEEN '2020-02-01' AND '2020-04-01')
                  AND SUM(t1.user_id >= '2020-02-01') > 1 

) DiffTable
WHERE diff IS NOT NULL

但是据说：

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

我该怎么办？

this is my fiddle

预期结果

+-----+-----+---------+
| MAX | MIN | AVERAGE |
+-----+-----+---------+
|  36 |   1 |      22 |
+-----+-----+---------+

解释：

- the user_id who have approval transaction on 2020-02-01 until 2020-04-01 and user_id who have transaction more than 1 are user_id 13 & 14
- the maximum of different day on 2020-02-01 until 2020-04-01 are user_id 13 which the different day for each transaction happen in 2020-03-04 and doing next transaction again in 2020-04-10
- the minimum day of different day of each transaction are user_id 14 who doing transaction on 2020-03-02 and next transaction 2020-03-03
- average are 22 days (sum of different day on user_Id 13 & 14 / amount of user_id who fit on this condition)

Answer 1

您需要在子查询之外执行 GROUPing；子查询应该只用于将 selected 事务限制为具有所需 status_id 值和所需范围内的日期的事务。那么就可以在外层查询select个时间段内有一笔以上交易的用户：

SELECT user_id,
       COUNT(*) AS transactions, 
       MIN(diff) AS `MIN`, 
       MAX(diff) AS `MAX`, 
       SUM(diff) / COUNT(diff) AS `AVG`
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)
    AND createdAt BETWEEN '2020-02-01' AND '2020-04-01'
) DiffTable
WHERE diff IS NOT NULL
GROUP BY user_id
HAVING COUNT(*) > 1

输出（对于你的fiddle）：

user_id     transactions    MIN     MAX     AVG
14          2               1       7       4.0000

Demo on dbfiddle

如果您想要基于该期间发生的所有笔交易的值，而不是 user_id，您可以简单地删除 GROUP BY 和HAVING 个子句：

SELECT COUNT(*) AS transactions, 
       MIN(diff) AS `MIN`, 
       MAX(diff) AS `MAX`, 
       SUM(diff) / COUNT(diff) AS `AVG`
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)
    AND createdAt BETWEEN '2020-02-01' AND '2020-04-01'
) DiffTable
WHERE diff IS NOT NULL

输出：

transactions    MIN     MAX     AVG
3               1       37      15.0000

Demo on dbfiddle

请注意，DATEDIFF 计算中的现有子查询存在一些问题：没有 ORDER BY 的 LIMIT 不能保证给出预期的结果，并且没有以 status_id 为条件。我已在更新的查询中修复了这两个问题。

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