Question

编辑：问题陈述：在3月初到5月底之间选择所有SIIIS和SULLS数据

对于SIIIS中的每个“XFFF”，查找SULLS中“XFFF”中的任何匹配，其中SULLS中的“Start_Time”位于SIIIS中的“Case_Create”之后，“Start_Time”和“Case_Create”之间的时差小于3小时。选择时间差最小的匹配

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

试试此代码。基本上它应该做的是它按DATEDIFF(hour, si.CASE_CREATE, su.START_TIME)按升序排序然后选择第一个1.它没有经过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

Answer 2

试图清理和修复你的代码，应该在sqlserver 2008+上运行，datediff不会给你几小时的差异（非常常见的误解），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

TSQL - 查找范围中的最小值

2 个答案: