我有一个如下所示的“订单”表:

+---------------+--------------+------------+
| customer_name | order_number |    date    |
+---------------+--------------+------------+
| jack          |            1 | 2018-01-01 |
| jack          |            2 | 2018-01-06 |
| jack          |            3 | 2018-01-19 |
| jack          |            4 | 2018-01-06 |
| jack          |            5 | 2018-02-27 |
| jack          |            6 | 2018-02-02 |
+---------------+--------------+------------+

现在，我想要一个表格，以天为单位给出连续日期的差异。像这样:

+------------+------------+------+
|    date    | next_date  | diff |
+------------+------------+------+
| 2018-01-01 | 2018-01-06 |    5 |
| 2018-01-06 | 2018-01-06 |    0 |
| 2018-01-06 | 2018-01-19 |   13 |
| 2018-01-19 | 2018-02-02 |   14 |
| 2018-02-02 | 2018-02-27 |   25 |
+------------+------------+------+

我使用的查询是这样的:

SELECT orders.date, MIN(table1.date) FROM orders
    LEFT JOIN orders table1
    on orders.customer_name = table1.customer_name
    AND table1.date >= orders.date
    AND table1.order_number !=  orders.order_number
    WHERE orders.customer_name = 'jack'
    GROUP BY orders.order_number, orders.date
    ORDER BY orders.date;

这是输出:

+------------+------------+
|    date    | next_date  |
+------------+------------+
| 2018-01-01 | 2018-01-06 |
| 2018-01-06 | 2018-01-06 |
| 2018-01-06 | 2018-01-06 |
| 2018-01-19 | 2018-02-02 |
| 2018-02-02 | 2018-02-27 |
| 2018-02-27 |    NULL    |
+------------+------------+

如您所见，存在一些问题。

1. 有两行的date和next_date都是2018-01-06。
2. 没有 next_date 为 2018-01-19 的行`
3. 最后一行的 next_date 的值为 NULL>
4. 如何获得日期的天数差异？

我知道这是因为我已经按 order_number 和 >= 分组了 但我不知道还有什么办法可以解决这个问题。我觉得有一个明显的简单解决方案正在逃避我。有帮助吗？

SQL Fiddle

如果 SQL Fiddle 不工作:

CREATE TABLE orders
    (`customer_name` varchar(4), `order_number` int, `date` varchar(10))
;

INSERT INTO orders
    (`customer_name`, `order_number`, `date`)
VALUES
    ('jack', 1, '2018-01-01'),
    ('jack', 2, '2018-01-06'),
    ('jack', 3, '2018-01-19'),
    ('jack', 4, '2018-01-06'),
    ('jack', 5, '2018-02-27'),
    ('jack', 6, '2018-02-02')
;

最佳答案

如果您对依赖于 MySQL 8.x 或更高版本的非 ROW_NUMBER() 解决方案感兴趣，请查看下一个说明。

解释:

1) 首先，我们从 orders 表中选择所有日期，通过升序对它们进行排序，并为每个日期分配一个虚拟的自动增量 ID。我们将得到这样的东西:

SELECT (@row_number := @row_number + 1) AS orderNum, date 
FROM ORDERS, (SELECT @row_number:=0) AS t
ORDER BY date;

Output:
1   2018-01-01
2   2018-01-06
3   2018-01-06
4   2018-01-19
5   2018-02-02
6   2018-02-27

2) 我们创建了一个与前一个类似的查询，但这次我们丢弃了第一行，如下所示:

SELECT (@row_number2 := @row_number2 + 1) AS orderNum, date
FROM ORDERS, (SELECT @row_number2 := 0) AS t
ORDER BY date
LIMIT 999999999999
OFFSET 1;

Output:
1   2018-01-06
2   2018-01-06
3   2018-01-19
4   2018-02-02
5   2018-02-27

这里唯一的问题是，我们必须将 LIMIT 数字硬编码为足够大的数字，这样我们才能确保我们会选择除第一行以外的所有行。

3) 此时，您应该考虑通过虚拟生成的 ID 连接之前的两个结果。那么，让我们看看最终的查询:

SELECT
    startDate.date AS date,
    nextDate.date AS next_date,
    DATEDIFF(nextDate.date, startDate.date) AS diff
FROM
    (SELECT (@row_number := @row_number + 1) AS orderNum, date 
     FROM ORDERS, (SELECT @row_number:=0) AS t
     ORDER BY date) AS startDate
INNER JOIN
    (SELECT (@row_number2 := @row_number2 + 1) AS orderNum, date
     FROM ORDERS, (SELECT @row_number2 := 0) AS t
     ORDER BY date
     LIMIT 999999999999
     OFFSET 1) AS nextDate ON nextDate.orderNum = startDate.orderNum;

Output:
2018-01-01  2018-01-06  5
2018-01-06  2018-01-06  0
2018-01-06  2018-01-19  13
2018-01-19  2018-02-02  14
2018-02-02  2018-02-27  25

您可以在此处查看工作示例:http://sqlfiddle.com/#!9/1572ea/27

关于mysql - 具有相同列的 MySQL 中的 DateDiff，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/52862381/