好的,这里有一个棘手的问题......如果我的数据如下所示:
表格1
ID Date_Created
1 1/1/2009
2 1/3/2009
3 1/5/2009
4 1/10/2009
5 1/15/2009
6 1/16/2009
如何获取相隔 2 天的记录?我的最终结果集应该是第 1-3 行和第 5-6 行。谢谢!
SELECT l.*
FROM Table1 l
INNER JOIN Table1 r ON DATEDIFF(d, l.Date_Created, r.Date_Created) = 2
AND r.Date_Created = (SELECT TOP 1 * FROM Table1 WHERE Date_Created > l.Date_Created ORDER BY Date_Create)
这行得通吗?
select t1.id, t2.id
from table1 t1
join table1 t2
on t2.date_created - t1.date_created <= 2
——这给了你什么?
选择 DISTINCT t1.id, t1.date_created, t2.id, t2.date_created from table1 t1, table1 t2 where datediff(dd,t1.date_created,t2.date_created) = 2 AND t1.id != t2.id ORDER BY t1 。ID;
我可能会建议使用编程代码来做到这一点。您想要收集行组(单独的组)。我认为您无法使用单个查询来解决此问题(这只会给您返回一组行)。
select distinct t1.*
from Table1 t1
inner join Table1 t2
on abs(cast(t1.Date_Created - t2.Date_Created as float)) between 1 and 2
如果你想得到相隔“N”天的行,你可以试试这个:
select t1.date_created, t2.date_created
from table1 t1, table1 t2
where t1.id <> t2.id and
t2.date_created-t1.date_created between 0 and N;
例如,正如您所说,如果您想获取 2 天内的行,您可以使用以下内容:
select t1.date_created,t2.date_created
from table1 t1, table1.t2
where t1.id <> t2.id and
t2.date_created-t1.date_created between 0 and 2;
我希望这有帮助....
问候,斯里克里希纳。
游标将是最快的,但这里有一个 SELECT 查询可以做到这一点。请注意,对于相隔“最多 N”天而不是 2 天,您必须将表二替换为从 0 到 N-1 的整数表(效率会变差)。
我承认这并不完全清楚你想要什么,但我猜你想要包含至少两行并且连续行最多相隔 2 天的行范围。如果日期与 ID 一起增加,这应该有效。
with Two as (
select 0 as offset union all select 1
), r2(ID, Date_Created_o, dr) as (
select
ID, Date_Created+offset,
Date_Created + offset - dense_rank() over (
order by Date_Created+offset
) from r cross join Two
)
select
min(ID) as start, max(ID) as finish
from r2
group by dr
having min(ID) < max(ID)
order by dr;