我有一个表'order'看起来像这样:
+---------------+--------------+------------+
| 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,我想要一张表,它给出了连续日期的差异。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。
date
和next_date
都是2018-01-06
。next_date
是2018-01-19`next_date
的最后一行有一个NULL值我知道这种情况正在发生,因为我按order_number和>=
分组,但我不知道如何处理这个问题。我觉得有一个明显的简单解决方案可以避开我。有帮助吗?
如果SQL Fiddle不起作用:
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')
;
如果您对非ROW_NUMBER()解决方案感兴趣,具体取决于MySQL 8.x或更高版本,请查看下一个解释。
说明:
1)首先,我们从订单表中选择所有日期,按升序排序并为每个日期分配虚拟自动增量ID。我们会得到这样的东西:
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)我们创建一个与前一个类似的查询,但这次我们丢弃第一行,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编号硬编码到足够高的数字,因此我们可以保证我们将选择除第一行之外的所有行。
3)此时,您应该考虑通过虚拟生成的ID加入之前的两个结果。所以,让我们看看最后的查询:
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
你可以在这里看到工作示例:http://sqlfiddle.com/#!9/1572ea/27
你可以这样做。但你的order 4
有previous date
而不是order 3
。所以它会产生负面影响。
SELECT customer_name,order_number,date,
LEAD(date) OVER (ORDER BY customer_name,order_number) next_date,
ISNULL(DATEDIFF(DAY,date,LEAD(date) OVER (ORDER BY customer_name,order_number)),0) AS diff
FROM #orders
Row_Number()
功能。orders
表两次用于两个不同的Derived Tables,行号根据date
的升序分配。两者之间的区别在于其中一个表将具有修改的行号(递增1)。DateDiff()
函数来确定日期之间的差异。请尝试以下方法:
SELECT t.`date`,
next_t.`date` AS next_date,
DATEDIFF(next_t.`date`, t.`date`) AS diff
FROM
(
SELECT 1 + (ROW_NUMBER() OVER (ORDER BY `date` ASC)) AS rn,
`date`
FROM orders
WHERE customer_name = 'jack'
) AS t
JOIN
(
SELECT (ROW_NUMBER() OVER (ORDER BY `date` ASC)) AS rn,
`date`
FROM orders
WHERE customer_name = 'jack'
) AS next_t ON next_t.rn = t.rn