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我有一张这样的桌子：

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);

这是我的小提琴

在该表中，有id每笔交易的 id、user_id用户、createdAt交易发生的日期以及每笔交易status_id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”之间进行交易的每个用户的每笔交易的最大、最小、平均不同的日期，并且在该期间批准了 >1 笔交易

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

但据说：

在没有 GROUP BY 的聚合查询中，SELECT 列表的表达式 #1 包含非聚合列 'fiddle_KDQIQDMUZEIOVXFHRZPY.t1.ID'；这与 sql_mode=only_full_group_by 不兼容

我应该怎么办？

这是我的小提琴

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)
4

1 回答 1

1

您需要在GROUP子查询之外进行 ing；子查询应仅用于将所选事务限制为具有所需status_id值和日期在所需范围内的事务。然后可以在外层查询中选择该期间内有超过一笔交易的用户：

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（为您的小提琴）：

user_id     transactions    MIN     MAX     AVG
14          2               1       7       4.0000

dbfiddle 上的演示

如果您想要基于该期间发生的所有user_id事务的值，而不是 by ，您可以简单地删除GROUP BYandHAVING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

dbfiddle 上的演示

请注意，DATEDIFF计算中的现有子查询存在几个问题： a LIMITwithoutORDER BY不能保证给出预期结果，并且没有条件 on status_id。我已经在更新的查询中解决了这两个问题。

于 2020-03-09T06:26:38.670 回答