SQL

Question

我有一个table喜欢

id | value | date     | type
----------------------------
1  | 2     | 1.1.2016 | 1
----------------------------
2  | 6     | 2.1.2016 | 1
----------------------------
3  | 1     | 7.1.2016 | 1
----------------------------
4  | 10    | 3.1.2016 | 2
----------------------------
5  | 8     | 8.1.2016 | 1

我需要编写一个查询，获取第 1 类条目，以便将它们之间没有第 2 类条目的条目汇总到同一行。在我的示例中，查询将 return

sum | start    | end
-------------------------
8   | 1.1.2016 | 2.1.2016
-------------------------
9   | 7.1.2016 | 8.1.2016

Answer 1

您可以通过计算行前非 1 值的数量来识别每个组。剩下的只是聚合：

select sum(value), min(date), max(date)
from (select t.*,
             sum(case when type <> 1 then 1 else 0 end) over (order by id) as grp
      from t
     ) t
where type = 1
group by grp;

Answer 2

您可以尝试 lead() 函数，如下面的查询

select 
      value+lead_value as value, 
      date as start_date, 
      end_date
from(
    select *, 
           lead(value) over(order by id asc) lead_value, 
           lead(date) over(order by id asc) end_date, 
           ROW_NUMBER() over(partition by t order by id asc) row_num 
    from t1
    where type = 1)t2
where row_num % 2 = 1;

这给了我以下结果

SQL - 根据条件求和到多行

SQL - sum to multiple rows based on condition

postgresql

gaps-and-islands