Question

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);

this is my fiddle

在该表中，每个交易的ID为id，用户为user_id，交易发生的日期为createdAt，状态为status_id每笔交易（在这种情况下，status_id 4、5、6、8是批准的交易）

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，平均不同日期，在该时期内批准了1个交易< / p>

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

但是有人说：

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

我该怎么办？

this is my fiddle

预期结果

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

说明：

- 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GROUP；子查询仅应用于将所选交易限制为具有所需status_id值和日期在所需范围内的交易。然后，您可以在外部查询中选择该期间内具有多个交易的用户：

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（用于您的小提琴）：

user_id     transactions    MIN     MAX     AVG
14          2               1       7       4.0000

Demo on dbfiddle

如果您希望基于该时期内发生的所有所有交易而不是user_id进行交易，则只需删除GROUP BY和HAVING条款：

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两个问题：没有LIMIT的{{1}}不能保证得到预期的结果，并且没有条件ORDER BY。我已经在更新的查询中解决了这两个问题。

如何使用mysql找出每个用户的每笔交易的时差（一天）

1 个答案: