MySQL window 函数获取该组状态相同的两个日期之间的差异

Question

我想做的事情：

查找 'awb' 的最后收到的行。获取该行中的 current_status。如果 current_status 是 'PICKUP EXCEPTION'、'OUT FOR PICKUP'、'PICKUP RESCHEDULED' 中的任何一个，则针对该特定 'awb' 找到这些状态首次出现的行检查'awb'的那些STATUSES第一次出现和最后一次出现之间的天数，并输出相差超过2天的'awbs'。

我做了什么：

WITH ranked_order_status AS (
  SELECT os.*,  
  ROW_NUMBER() OVER (PARTITION BY awb ORDER BY STR_TO_DATE(recived_at,'%m/%d/%y %T') desc) AS occurance
  FROM order_status AS os 
),
filtered_last_occurance_in_list as (
 select * from ranked_order_status where occurance=1 and current_status in ('PICKUP EXCEPTION', 'OUT FOR PICKUP', 'PICKUP RESCHEDULED')
),
all_awb_with_last_status_in_list as 
(
 select * from order_status os2  where awb in (select awb from filtered_last_occurance_in_list) order by awb
)
select id, recived_at, awb,current_status from all_awb_with_last_status_in_list

当前输出：

id  |recived_at   |awb           |current_status    |
----|-------------|--------------|------------------|
2413|5/7/21 6:13  | 1091108643160|OUT FOR PICKUP    |
2414|5/7/21 6:14  | 1091108643160|OUT FOR PICKUP    |
2501|5/7/21 10:55 | 1091108643160|PICKUP EXCEPTION  |
2503|5/7/21 11:13 | 1091108643160|PICKUP EXCEPTION  |
1337|4/30/21 7:44 |14325421720410|OUT FOR PICKUP    |
1399|4/30/21 17:39|14325421720410|PICKUP EXCEPTION  |
1476|5/1/21 6:09  |14325421720410|PICKUP RESCHEDULED|
1500|5/1/21 11:09 |14325421720410|PICKUP EXCEPTION  |
  77|4/22/21 7:18 |19041127718646|OUT FOR PICKUP    |
 121|4/22/21 11:17|19041127718646|PICKUP EXCEPTION  |
  81|4/22/21 7:18 |19041128001310|OUT FOR PICKUP    |
 124|4/22/21 11:18|19041128001310|PICKUP EXCEPTION  |
  83|4/22/21 7:18 |19041128018585|OUT FOR PICKUP    |
 125|4/22/21 11:18|19041128018585|PICKUP EXCEPTION  |
 979|4/27/21 12:19|19041129734005|PICKUP EXCEPTION  |
1334|4/30/21 6:48 |19041130453211|OUT FOR PICKUP    |
1372|4/30/21 11:19|19041130453211|PICKUP EXCEPTION  |
1534|5/1/21 18:47 |19041130453211|PICKUP EXCEPTION  |
1483|5/1/21 7:18  |19041130642631|OUT FOR PICKUP    |
1490|5/1/21 8:18  |19041130642631|PICKUP EXCEPTION  |
1688|5/3/21 10:47 |19041130642631|PICKUP EXCEPTION  |

预期结果： 假设 awb = 1091108643160，我需要使用 (5/7/21 11:13 - 5 创建一个名为 delay 的新列/7/21 10:55) = 18m

 id   recived_at      awb           current_status 
2413|5/7/21 6:13  | 1091108643160|OUT FOR PICKUP    |
2414|5/7/21 6:14  | 1091108643160|OUT FOR PICKUP    |
2501|5/7/21 10:55 | 1091108643160|PICKUP EXCEPTION  | -> this is first occurrence of same status
2503|5/7/21 11:13 | 1091108643160|PICKUP EXCEPTION  | -> this is last row

Answer 1

calculate date difference between last row of a awb and first occurrence of that (status and awb)

WITH cte AS ( SELECT DISTINCT
                     awb, 
                     FIRST_VALUE(current_status) OVER (PARTITION BY awb ORDER BY date DESC) current_status
              FROM source )
SELECT awb, DATEDIFF(MAX(date), MIN(date))
FROM source
JOIN cte USING (awb, current_status)
GROUP BY awb

MySQL window 函数获取该组状态相同的两个日期之间的差异

MySQL window function get difference between two dates where status is same on that group

mysql

window-functions