mysql中具有相同列的datediff

yshpjwxd  于 2021-06-20  发布在  Mysql
关注(0)|答案(3)|浏览(349)

我有一张“订单”表,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,我想要一个表格,它给出连续日期的差值(以天为单位)。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:

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    |
+------------+------------+

正如你所看到的,有几个问题。
有两排 date 以及 next_date 两者都是 2018-01-06 .
没有一排 next_date is 2018-01-19最后一行的值为空next_date我怎样才能得到以天为单位的日期差异? 我知道这是因为我已经按订单号和>=` 但我不知道还有什么办法。我觉得有一个显而易见的简单的解决办法,就是逃避我。有什么帮助吗?
sql小提琴
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')
;
dtcbnfnu

dtcbnfnu1#

你可以这样做。但是你的 order 4 有一个 previous dateorder 3 . 所以它会产生一个负值。

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
gtlvzcf8

gtlvzcf82#

如果您对基于mysql 8.x或更高版本的非row\u number()解决方案感兴趣,请查看下一个解释。
说明:
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递增et id。我们会得到这样的结果:

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,但这次我们放弃了第一行,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,我们必须将限制数字硬编码为足够高的数字,这样我们可以确保我们将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连接前面的两个结果。因此,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

您可以在这里看到工作示例:http://sqlfiddle.com/#!1572ea/27年9月

yk9xbfzb

yk9xbfzb3#

订单号和日期之间没有直接关系。订单号越高,日期越短。
所以,我们不能用订单号来决定下一个订单日期。
由于您的mysql版本是8.0,我们可以对 Row_Number() 功能。
我们将使用 orders 表两次放入两个不同的派生表中,行号是根据表的升序指定的 date . 两者之间的区别在于其中一个表将修改行号(递增1)。
现在,我们需要做的就是根据行号值在它们之间进行内部连接。由于行号之间的“滑动间隔”,我们将得到“当前”和“下一个”日期。内部连接还将确保没有匹配“next”日期的最后一个“current”日期行不会出现。
最终,我们可以使用 DateDiff() 函数来确定日期之间的差异。
请尝试以下操作:

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

db小提琴演示

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