MySQL - 找出ROW-LAG和LEAD之间的时差不起作用

Question

我在图片screenshot of my table

上有一张表格

customer_id   purchase_date     Difference between purchases (in days) 
                                     = WHAT I AM TRYING TO GET
1             23/04/2017        0 (first )row
1             24/04/2017        1
1             01/01/2018        252
2             03/05/2017        0 (this is a new customer)
2             10/05/2017        7

我想计算同一客户的两次购买之间的时差。我尝试使用LAG和LEAD函数，但是我得到了语法错误，我不明白

现在我一直这样做：

SELECT customer_id, purchase_date
case 
when lag(purchase_date,1,0) over (partition by customer_id order by 
purchase_date) = 0 then 0
ELSE purchase_date -lag(purchase_date,1,0) over(partition by customer_id order 
by purchase_date) 
end 
FROM
Table1

它给我一个语法错误，我在第一个＆＃34; over＆＃34;

之后才明白

Answer 1

当前的MySQL版本不支持LAG和LEAD等窗口函数。
MySQL 8.0+现在是候选版本，它将支持窗口函数，但不支持生产就绪。

在当前的MySQL版本中，您可以使用MySQL的用户变量或与共同相关的子查询来模拟LAG。

创建表/插入数据

CREATE TABLE Table1
    (`customer_id` int, `purchase_date` varchar(10))
;

INSERT INTO Table1
    (`customer_id`, `purchase_date`)
VALUES
    (1, '23/04/2017'),
    (1, '24/04/2017'),
    (1, '01/01/2018'),
    (2, '03/05/2017'),
    (2, '10/05/2017')
;

MySQL用户变量的技巧是正确初始化它们。

<强>查询

SELECT 
 *
 , (@customer_id := Table1.customer_id) AS init_customer_id_param
 , (@purchase_date := Table1.purchase_date) AS init_purchase_date_param
FROM 
 Table1
CROSS JOIN (
 SELECT
      @customer_id := NULL
   ,  @purchase_date := NULL 
)
 AS init_user_params

<强>结果

| customer_id | purchase_date | @customer_id := NULL | @purchase_date := NULL | init_customer_id_param | init_purchase_date_param |
|-------------|---------------|----------------------|------------------------|------------------------|--------------------------|
|           1 |    23/04/2017 |               (null) |                 (null) |                      1 |               23/04/2017 |
|           1 |    24/04/2017 |               (null) |                 (null) |                      1 |               24/04/2017 |
|           1 |    01/01/2018 |               (null) |                 (null) |                      1 |               01/01/2018 |
|           2 |    03/05/2017 |               (null) |                 (null) |                      2 |               03/05/2017 |
|           2 |    10/05/2017 |               (null) |                 (null) |                      2 |               10/05/2017 |

现在您可以添加计算部分。
请记住，在MySQL的用户变量初始化之前，需要完成计算。
所以MySQL用户变量具有前一列值。

<强>查询

SELECT 
 *
 , (
     CASE
       WHEN (@customer_id = Table1.customer_id)
       THEN DATEDIFF(STR_TO_DATE(purchase_date, "%d/%m/%Y"), STR_TO_DATE(@purchase_date, "%d/%m/%Y"))
     END
   ) AS diff
 , (@customer_id := Table1.customer_id) AS init_customer_id_param
 , (@purchase_date := Table1.purchase_date) AS init_purchase_date_param   
FROM 
 Table1
CROSS JOIN (
 SELECT
      @customer_id := NULL
   ,  @purchase_date := NULL 
)
 AS init_user_params
ORDER BY 
  STR_TO_DATE(purchase_date, "%d/%m/%Y") ASC

注意

我使用STR_TO_DATE函数将基于varchar的dateformat格式化为datetime格式。
如果您的日期列已经是日期数据类型，则可以删除该函数并改为使用THEN DATEDIFF(purchase_date, @purchase_date)。

<强>结果

| customer_id | purchase_date | @customer_id := NULL | @purchase_date := NULL |   diff | init_customer_id_param | init_purchase_date_param |
|-------------|---------------|----------------------|------------------------|--------|------------------------|--------------------------|
|           1 |    23/04/2017 |               (null) |                 (null) | (null) |                      1 |               23/04/2017 |
|           1 |    24/04/2017 |               (null) |                 (null) |      1 |                      1 |               24/04/2017 |
|           1 |    01/01/2018 |               (null) |                 (null) |    252 |                      1 |               01/01/2018 |
|           2 |    03/05/2017 |               (null) |                 (null) | (null) |                      2 |               03/05/2017 |
|           2 |    10/05/2017 |               (null) |                 (null) |      7 |                      2 |               10/05/2017 |

现在只需选择正确的列。

<强>查询

SELECT 
   Table1_user_params.customer_id 
 , Table1_user_params.purchase_date
 , (
     CASE
       WHEN  Table1_user_params.diff IS NULL
       THEN 0
       ELSE Table1_user_params.diff 
     END 
   ) 
    AS diff
FROM ( 
  SELECT 
   *
   , (
       CASE
         WHEN (@customer_id = Table1.customer_id)
         THEN DATEDIFF(STR_TO_DATE(purchase_date, "%d/%m/%Y"), STR_TO_DATE(@purchase_date, "%d/%m/%Y"))
       END
     ) AS diff
   , (@customer_id := Table1.customer_id) AS init_customer_id_param
   , (@purchase_date := Table1.purchase_date) AS init_purchase_date_param   
  FROM 
   Table1
  CROSS JOIN (
    SELECT
        @customer_id := NULL
     ,  @purchase_date := NULL 
  )
   AS init_user_params
  ORDER BY 
    STR_TO_DATE(purchase_date, "%d/%m/%Y") ASC
) 
 AS Table1_user_params

<强>结果

| customer_id | purchase_date | diff |
|-------------|---------------|------|
|           1 |    23/04/2017 |    0 |
|           1 |    24/04/2017 |    1 |
|           1 |    01/01/2018 |  252 |
|           2 |    03/05/2017 |    0 |
|           2 |    10/05/2017 |    7 |