TSQL - 查找范围中的最小

Question

编辑： 问题陈述： 在SIIIS选择三月开始和结束的五月

之间所有SIIIS和SULLS数据对于每个“XFFF”寻找在“XFFF任何匹配“在SULLS中，SULLS中的”Start_Time“在SIIIS中的”Case_Create“之后，并且”Start_Time“和”Case_Create“之间的时间差小于3小时。 选择最小时差的比赛

select 
si.XFFF xfff,  
DATEDIFF(hour, si.CASE_CREATE, su.START_TIME) diffInhours 
from siiis si inner join sulls su on si.xfff= su.xfff 
where 
((si.CASE_CREATE BETWEEN '20130301' AND '20130531') and (su.START_TIME between '20130301'AND '20130531')) 
and 
(su.START_TIME > si.CASE_CREATE) and 
(DATEDIFF(hour, si.CASE_CREATE, su.START_TIME) < 3 and DATEDIFF(hour, si.CASE_CREATE, su.START_TIME) > 0) 

Answer 1

试图清理和修复你的代码，应该SQLSERVER 2008+运行，DATEDIFF不会给你时间差（很常见的误解），所以我采取了不同的方法：

select 
si.num_phone Phone, 
su.start_time sulstarrtTime , 
si.CASE_CREATE siStartTime, 
datediff(hour, 0, su.START_TIME - si.CASE_CREATE) diffInhours 
from siiis si 
cross apply 
(select top 1 * from sulls su 
where 
si.xfff= su.xfff 
and su.START_TIME > si.CASE_CREATE 
-- less than 3 hours difference 
and dateadd(hour, 3, si.CASE_CREATE) > su.START_TIME 
-- more than 1 hour difference your code indicated you needed this, your text said otherwise 
--and dateadd(hour, 1, si.CASE_CREATE) <= su.START_TIME 
order by START_TIME 
) su 
WHERE si.CASE_CREATE BETWEEN '20130301' AND '20130531' -- you are not including all of may 

Answer 2

试试这段代码。基本上它应该做的是它按照DATEDIFF(hour, si.CASE_CREATE, su.START_TIME)的顺序从小到大排列，然后选择第一个。它没有测试，所以我解释了逻辑。

SELECT TOP 1 
si.num_phone Phone, 
su.start_time sulstarrtTime , 
si.CASE_CREATE siStartTime, 
DATEDIFF(hour, si.CASE_CREATE, su.START_TIME) diffInhours 
from siiis si inner join sulls su on si.num_phone = su.num_phone 
where 
((si.CASE_CREATE BETWEEN '20130301' AND '20130531') and (su.START_TIME between '20130301'AND '20130531')) 
and 
(su.START_TIME > si.CASE_CREATE) and 
(DATEDIFF(hour, si.CASE_CREATE, su.START_TIME) < 3 and DATEDIFF(hour, si.CASE_CREATE, su.START_TIME) > 0) 
order by DATEDIFF(hour, si.CASE_CREATE, su.START_TIME) ASC