获取此日期列并将它们分组 1-30、31-59、60-89、+90，包括即将到来的日期和已经过去的日期

Question

enter image description here我有一个日期列，我试图将其分组到特定的组中。一些日期已经过去，其他日期在未来。我想按 1-30、31-59、60-89、+90 对它们进行分组，因此已经通过的和即将通过的总共有 8 个组。以下是我到目前为止所写的内容，但我觉得我过于复杂了，我做的调整越多，它就变得越不正确。感谢任何见解！

CASE WHEN DATEDIFF(CURRENT_DATE(),`date`) >= 90 THEN '90 days past'
     WHEN DATEDIFF(CURRENT_DATE(),`date`) >= 60 THEN '60 days past'
     WHEN DATEDIFF(CURRENT_DATE(),`date`) >= 30 THEN '30 days past'
     WHEN DATEDIFF(`date`,CURRENT_DATE()) >= 90 THEN '90 days future'
     WHEN DATEDIFF(`date`,CURRENT_DATE()) >= 60 THEN '60 days future'
     WHEN DATEDIFF(`date`,CURRENT_DATE()) >= 30 THEN 'Next 30 days'
     ELSE 'Not Late'
END

Answer 1

我觉得还行

我会为每个使用相同的 datediff。

create table test (`date` date);
insert into test values
( date_add(current_date, interval -90 day) ), 
( date_add(current_date, interval -89 day) ), 
( date_add(current_date, interval -60 day) ), 
( date_add(current_date, interval -59 day) ), 
( date_add(current_date, interval -30 day) ), 
( date_add(current_date, interval -29 day) ),
( date_add(current_date, interval 29 day) ),
( date_add(current_date, interval 30 day) ), 
( date_add(current_date, interval 59 day) ), 
( date_add(current_date, interval 60 day) ), 
( date_add(current_date, interval 89 day) ), 
( date_add(current_date, interval 90 day) )

select `date`, 
 CASE
 WHEN DATEDIFF(`date`, CURRENT_DATE) <= -90 THEN '90 days past'
 WHEN DATEDIFF(`date`, CURRENT_DATE) <= -60 THEN '60 days past'
 WHEN DATEDIFF(`date`, CURRENT_DATE) <= -30 THEN '30 days past'
 WHEN DATEDIFF(`date`, CURRENT_DATE) >=  90 THEN '90 days future'
 WHEN DATEDIFF(`date`, CURRENT_DATE) >=  60 THEN '60 days future'
 WHEN DATEDIFF(`date`, CURRENT_DATE) >=  30 THEN 'Next 30 days'
 ELSE 'Not Late'
 END as status
from test

date       | status        
:--------- | :-------------
2021-10-12 | 90 days past  
2021-10-13 | 60 days past  
2021-11-11 | 60 days past  
2021-11-12 | 30 days past  
2021-12-11 | 30 days past  
2021-12-12 | Not Late      
2022-02-08 | Not Late      
2022-02-09 | Next 30 days  
2022-03-10 | Next 30 days  
2022-03-11 | 60 days future
2022-04-09 | 60 days future
2022-04-10 | 90 days future

db<>fiddle here

获取此日期列并将它们分组 1-30、31-59、60-89、+90，包括即将到来的日期和已经过去的日期

Taking this column of dates and grouping them 1-30, 31- 59, 60 - 89, +90, both upcoming and those dates that have passes

sql

grouping

date