Question

我遇到了一部分查询，无法提取日期差异至少为三个月的row_numbers。所以在下面的例子中，我想提取row_number 1（总是第一个），5和6.所以在row_number 1之后row_numbers有一个date_diff＆gt; 3个月（并且在第二次提取row_number之后再次应用此条件，直到无。在SQL中是否有任何允许这样的条件的函数或方法？

table_name: users  

id  row_number  User    date          
---|----------|-------|---------------------|
1  |1         | Usr1  | 2017-10-01 12:35:00 | 
2  |2         | Usr1  | 2017-10-01 12:35:00 | 
3  |3         | Usr1  | 2017-12-03 07:47:00 |
4  |4         | Usr1  | 2018-01-10 07:47:00 | 
5  |5         | Usr1  | 2018-02-10 07:47:00 | 
6  |6         | Usr1  | 2018-04-10 07:47:00 |

Answer 1

您可以使用lag()函数计算差异：

select *
from (
  select id, row_number, "User", date, 
         date - lag(date) over (order by id) as diff
  from users
) t
where diff is null  -- first row
   or diff > interval '3 month';

Answer 2

我不确定你想将间隔比较为几个月。我认为它们通常表示为几天。

所以，我会这样说：

select u.*
from (select u.*, lag(date) over (order by id) as prev_date
      from users u
     )  u
where prev_date is null or prev_date < date - interval '3 month';

如果or困扰您，您可以使用lag()中的默认值将其删除：

select u.*
from (select u.*, lag(date, 1, date - interval '100 year') over (order by id) as prev_date
      from users u
     )  u
where prev_date < date - interval '3 month';

如何在sql中获取row_numbers，其中condition是特定的日期差异？

2 个答案: