如何仅使用最后一个事务来计算 timediff mysql 5.7

Question

这是来自find out time difference for every user in condition mysql 5.7的继续问题

这是我的小提琴 https://dbfiddle.uk/?rdbms=mysql_5.7&fiddle=31b3be9d1e2444eb0b32c262176aa4b4

我有这张桌子

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

INSERT INTO test VALUES
  (1, 13, '2020-01-01', 8),
  (2, 13, '2020-01-03', 8),
  (3, 13, '2020-01-06', 8),
  (4, 13, '2020-01-02', 7),
  (5, 13, '2020-01-03', 7),
  (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 where status_id != 7
order by createdAt;


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

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“2020-02-01”到“2020-04-01”时间范围内每个重复用户的每个批准交易的时间差，重复用户是正在做的用户在时间范围结束之前的事务，并且在该时间范围内至少再次进行了 1 次事务，在这种情况下，用户在 '2020-04-01' 之前进行了批准事务，并且在这之间至少再次进行了 1 次批准事务2020-02-01' 和 '2020-04-01'。

对于那个问题，我根据@Akina 的回答使用了这个查询

-- Get pairs (current transaction, previous transaction) for these users

SELECT t1.user_id, 
       t1.createdAt, 
       t2.createdAt,
       DATEDIFF(t2.createdAt, t1.createdAt) diff
-- table for a transaction
FROM test t1
-- table for prev. transaction
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)
-- get data only for users from prev. query
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') > 1
         AND SUM(t3.createdAt BETWEEN '2020-02-01' AND '2020-04-01')) t4 ON t1.user_id = t4.user_id
-- check that there is no approved transaction between selected transactions
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)

the output table was like this 
+----------+------------+------------+------+
|  user_id | createdAt  | createdAt  | diff |
+----------+------------+------------+------+
|       13 | 2020-01-01 | 2020-01-03 |    2 |
|       13 | 2020-01-03 | 2020-01-06 |    3 |
|       14 | 2020-03-02 | 2020-03-03 |    1 |
|       13 | 2020-01-06 | 2020-03-04 |   58 |
|       14 | 2020-03-03 | 2020-03-10 |    7 |
+----------+------------+------------+------+

问题是，这个查询为每个用户计算时间范围内的时间差（'2020-02-01' 到 '2020-04-01'），并且计算时间范围之前的时间差（参见 users_id 13，用户还计算日期“2020-01-01”到“2020-01-03”的时间差）。我想要的是，如果用户在时间范围之前有交易，我只想计算他的 users_id 在时间范围之前的最后一次交易（在这种情况下，users_id 13 我只想计算'2020-01-06'中的时间差直到 '2020-03-04'，因为 2020 年 1 月 6 日是用户在时间范围之前最后一次交易的日期。这样的话，预期的结果是这样的：

+---------+------------+------------+------+
| user_id | createdAt  | createdAt  | diff |
+---------+------------+------------+------+
|      14 | 2020-03-02 | 2020-03-03 |    1 |
|      13 | 2020-01-06 | 2020-03-04 |   58 |
|      14 | 2020-03-03 | 2020-03-10 |    7 |
+---------+------------+------------+------+

Answer 1

我认为您只需排除在时间框架BETWEEN '2020-02-01' AND '2020-04-01 开始之前结束的每笔交易，因为它们不感兴趣。 因为您已经排除了超出时间范围的所有内容

您仍然知道何时在事务开始和结束时输入数据，因此您应该在额外的列中标记属于的行，这将使您的查询更简单，除了您的状态不可用，因为它会重复自己

SELECT t1.user_id, 
       t1.createdAt, 
       t2.createdAt createcompare,
       DATEDIFF(t2.createdAt, t1.createdAt) diff
-- table for a transaction
FROM test t1
-- table for prev. transaction
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') > 1
         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) 
HAViNG createcompare > '2020-02-01'

用户 ID |创建于 |比较 |差异 ------: | :--------- | :----------- | ---: 14 | 2020-03-02 | 2020-03-03 | 1 13 | 2020-01-06 | 2020-03-04 | 58 14 | 2020-03-03 | 2020-03-10 | 7 13 | 2020-03-04 | 2020-04-10 | 37

db小提琴here

更新：

这实际上更有意义了

SELECT t1.user_id, 
       t1.createdAt cretecompare1, 
       t2.createdAt cretecompare2,
       DATEDIFF(t2.createdAt, t1.createdAt) diff
-- table for a transaction
FROM test t1
-- table for prev. transaction
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') > 1
         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) 
HAViNG cretecompare2  BETWEEN '2020-02-01' AND '2020-04-01'

用户 ID |比较1 |比较2 |差异 ------: | :------------ | :------------ | ---: 14 | 2020-03-02 | 2020-03-03 | 1 13 | 2020-01-06 | 2020-03-04 | 58 14 | 2020-03-03 | 2020-03-10 | 7

db小提琴here