Question

我有一个表，用于在目录中注册产品价格的历史记录（另一个表）。

任何产品的价格都可以随时更改，因此我只需要确定上次更改价格的产品。例如，我有此表：

+----+--------------+-------+---------------------+
| id | product_code | price | price_date          |
+----+--------------+-------+---------------------+
|  1 | P01          | 10.00 | 2019-01-01 00:00:00 |
|  2 | P01          | 15.00 | 2019-01-03 00:00:00 |
|  3 | P01          | 20.00 | 2019-01-05 00:00:00 |
|  4 | P01          | 15.00 | 2019-01-09 00:00:00 |
|  5 | P02          | 10.00 | 2019-01-04 00:00:00 |
|  6 | P02          | 15.00 | 2019-01-06 00:00:00 |
|  7 | P02          | 15.00 | 2019-01-07 00:00:00 |
|  8 | P03          | 15.00 | 2019-01-09 00:00:00 |
|  9 | P04          | 15.00 | 2019-01-01 00:00:00 |
| 10 | P04          | 15.00 | 2019-01-02 00:00:00 |
| 11 | P04          | 25.00 | 2019-01-05 00:00:00 |
| 12 | P05          | 15.00 | 2019-01-01 00:00:00 |
| 13 | P05          | 15.00 | 2019-01-02 00:00:00 |
+----+--------------+-------+---------------------+

我需要这个结果（如果它们相同，我需要每个产品的倒数第二个和最后一个价格，如果它们相同，则忽略该产品）：

+--------------+-------+---------------------+
| product_code | price | price_date          |
+--------------+-------+---------------------+
|          P01 | 20.00 | 2019-01-05 00:00:00 |
|          P01 | 15.00 | 2019-01-09 00:00:00 |
|          P04 | 15.00 | 2019-01-02 00:00:00 |
|          P04 | 25.00 | 2019-01-05 00:00:00 |
+--------------+-------+---------------------+

这是例外：

P02并未更改价格，因此我将其忽略（是的，同一价格可以连续多次注册）

P03只有1个寄存器，因此从技术上讲它没有更改价格，因此我将其忽略（这非常少见，但某些产品从未更改价格）

P05没改变价格，所以我忽略了它

我放置这些数据是为了便于阅读，但从技术上讲，数据可以是任何顺序。

我向您发送查询，这样您就不会浪费太多时间：

CREATE TABLE PriceHistory(
id BIGINT UNSIGNED AUTO_INCREMENT NOT NULL,
product_code CHAR(3) NOT NULL,
price DECIMAL(5,2) NOT NULL,
price_date DATETIME NOT NULL,
PRIMARY KEY (id));

INSERT INTO PriceHistory(product_code, price, price_date) VALUES
('P01', 10, '2019-01-01'),
('P01', 15, '2019-01-03'),
('P01', 20, '2019-01-05'),
('P01', 15, '2019-01-09'),
('P02', 10, '2019-01-04'),
('P02', 15, '2019-01-06'),
('P02', 15, '2019-01-07'),
('P03', 15, '2019-01-09'),
('P04', 15, '2019-01-01'),
('P04', 15, '2019-01-02'),
('P04', 25, '2019-01-05'),
('P05', 15, '2019-01-01'),
('P05', 15, '2019-01-02');

谢谢。

Answer 1

这不是精确的答案，因为我希望每行都有结果，但是这个答案足够好并且可以完成工作：

SELECT info1.*, info2.* FROM
(SELECT lasts.product_code, lasts.id AS last_id, penultimate.id AS penultimate_id FROM
    (SELECT product_code, MAX(id) AS id FROM PriceHistory GROUP BY product_code HAVING COUNT(product_code) > 1) lasts
INNER JOIN
    (SELECT product_code, MAX(id) AS id FROM PriceHistory WHERE id NOT IN (SELECT MAX(id) FROM PriceHistory GROUP BY product_code) GROUP BY product_code) penultimate
ON lasts.product_code = penultimate.product_code) relations
INNER JOIN PriceHistory info1 ON relations.last_id = info1.id
INNER JOIN PriceHistory info2 ON relations.penultimate_id = info2.id
WHERE info1.price != info2.price

并给出以下结果：

+----+--------------+-------+---------------------+----+--------------+-------+---------------------+
| id | product_code | price | price_date          | id | product_code | price | price_date          |
+----+--------------+-------+---------------------+----+--------------+-------+---------------------+
|  4 | P01          | 15.00 | 2019-01-09 00:00:00 |  3 | P01          | 20.00 | 2019-01-05 00:00:00 |
| 11 | P04          | 25.00 | 2019-01-05 00:00:00 | 10 | P04          | 15.00 | 2019-01-02 00:00:00 |
+----+--------------+-------+---------------------+----+--------------+-------+---------------------+

选择更改其上一个值的记录

1 个答案: