SQL 编写显示日期之间跳过的查询

Question

我尝试编写一个查询，通过显示跳跃开始前的日期和跳过后的第一个日期来查找跳跃。

我试过了：

SELECT OrderDate FROM Orders

我得到了：

但我没有成功地将它们分成这样的空隙：

我该怎么做？

Answer 1

使用可以使用dense_rank()生成一个序列。当您从日期中减去它时，您将识别相邻的连续日期。

所以：

select min(orderdate), max(orderdate)
from (select o.*, dense_rank() over (order by orderdate) as seqnum
      from orders o
     ) o
group by dateadd(day, -seqnum, orderdate)
order by min(orderdate);

您需要 dense_rank()，因为您的数据有重复项。

编辑：

要查找跳跃，请使用 LEAD():

select dateadd(day, 1, orderdate),
       dateadd(day, -1, next_orderdate)
from (select orderdate, lead(orderdate) over (order by orderdate) as next_orderdate
      from (select distinct orderdate orders o) o
     ) o
where next_orderdate > dateadd(day, 1, orderdate);

Answer 2

试试这个：

with cte as (select ROW_NUMBER() OVER(ORDER BY OrderDate) as row_num,OrderDate from orders)
select 
curr.OrderDate as Gap_start, 
nex.OrderDate as Gap_end
from 
cte curr
left join
cte nex
on
curr.row_num = nex.row_num - 1;

SQL 编写显示日期之间跳过的查询

SQL write query that show the skipping between dates

sql

sql-server

northwind