1. sql 查找连续出现重复记录。。。是连续的哦
给你说说思想已经很好了
还要人给你代码。
sql:sleect 机台ID,物料ID,结论ID from table where 结论='不合格' group 机台ID,物料ID having
sum(结论)>=3
2. SQL语句如何选择3条连续的记录
如果ID值都是连续的话WHERE ID>=X AND ID<X+3
3. sqlserver中如何查询出连续日期记录的代码
你这个需要借助一个流水表，然后通过指定日期对流水表，进行数据叠加操作，
类似流水日期生成方法
---生成4月份所有的天数
selectdateadd(day,number,'2018-4-1')asdfrommaster..spt_values
wheretype='p'
andnumber>=0
anddateadd(day,number,'2018-4-1')between'2018-4-1'anddateadd(day,-1,'2018-5-1')
请试一试，如有疑问，及时沟通！
4. SQL筛选日期连续的记录
select empid as 员工编号,workdate as 旷工日期一,dateadd(day,1,workdate) as 旷工日期二 from table t1
where
exists
(select * from table where empid=t1.empid and datediff(day,t1.workdate,workdate)=1)
and not exists
(select * from table where empid=t1.empid and datediff(day,t1.workdate,workdate)=2)
and not exists
(select * from table where empid=t1.empid and datediff(day,t1.workdate,workdate)=-1)
存在大一天的数，但不存在大两天的数，也不存在小一天的数，就是正确结果了
5. sqlserver中如何查询出连续日期记录的代码
你这个需要借助一个流水表，然后通过指定日期对流水表，进行数据叠加操作，
类似流水日期生成方法
---生成4月份所有的天数
select dateadd(day,number,'2018-4-1') as d from master..spt_values
where type='p'
and number >=0
and dateadd(day,number,'2018-4-1') between '2018-4-1' and dateadd(day,-1,'2018-5-1')
请试一试，如有疑问，及时沟通！
6. SQL数据库中查询连续编号的的数据。
DECLARE@T1table(UserIDint,[name]nvarchar(50),numint);
insertinto@T1(UserID,[name],num)values(1001,'a',8)
insertinto@T1(UserID,[name],num)values(1002,'b',6)
insertinto@T1(UserID,[name],num)values(1003,'c',8)
insertinto@T1(UserID,[name],num)values(1004,'a',8)
insertinto@T1(UserID,[name],num)values(1005,'b',8)
select*from@t1
selecta.*from
(selectUserID,[name],[num]from@t1)ajoin
(selectUserID,[name],[num]from@t1)bona.UserID<b.UserIDanda.UserID+1=b.UserID
anda.num=b.num
groupbya.userid,a.[name],a.[num]
得到结果:1003c8
1004a8
7. sql server表中 怎么查询某些连续的记录，比如：查询第100到150行，表是无序排列的。
select top 150 * from 表的名字
except
select top 100 * from 表的名字
8. sql查询连续出勤记录
--用的好给加点分哦 特意帮你写的 sql2000 通过
declare @tab table (dt datetime , ur varchar(12))
insert into @tab
select '2007-01-02','2014' union all
select '2007-01-04','2014' union all
select '2007-01-08','2014' union all
select '2007-01-09','2014' union all
select '2007-01-10','2014' union all
select '2007-01-11','2014' union all
select '2007-01-13','2014' union all
select '2007-01-14','2014' union all
select '2007-01-15','2014' union all
select '2007-01-17','2014' union all
select '2007-01-18','2014' union all
select '2007-01-20','2014' union all
select '2007-01-21','2014' union all
select '2007-01-22','2014' union all
select '2007-01-23','2014' union all
select '2007-01-25','2014' union all
select '2007-01-29','2014'
select * from @tab --显示模拟数据
declare @tabOut table ([id] int,dtBegin datetime , dtEnd datetime)
declare @begin datetime,@end datetime,@n int
set @begin='1900-01-01'
set @n=0
declare my_cursor cursor for
select dt from @tab order by dt
Open my_cursor
Fetch Next From my_cursor into @end
while @@fetch_status=0
begin
if(datediff(day,@begin,@end)<>1)
begin
update @tabOut set dtEnd=@begin where [id]=@n
set @n=@n+1
insert into @tabOut([id],dtBegin)values(@n,@end)
end
set @begin=@end
Fetch Next From my_cursor into @end
end
update @tabOut set dtEnd=@begin where [id]=@n
close my_cursor
deallocate my_cursor
select * from @tabOut --结果
9. sql查询连续出勤记录
楼主的笔误很多啊，看看是不是这个效果。
declare @tab table (dt datetime , ur varchar(12))
insert into @tab
select '2007-01-02','2014' union all
select '2007-01-04','2014' union all
select '2007-01-08','2014' union all
select '2007-01-09','2014' union all
select '2007-01-10','2014' union all
select '2007-01-11','2014' union all
select '2007-01-13','2014' union all
select '2007-01-14','2014' union all
select '2007-01-15','2014' union all
select '2007-01-17','2014' union all
select '2007-01-18','2014' union all
select '2007-01-20','2014' union all
select '2007-01-21','2014' union all
select '2007-01-22','2014' union all
select '2007-01-23','2014' union all
select '2007-01-25','2014' union all
select '2007-01-29','2014'
select *
from @tab a
where exists(select * from @tab where datediff(dd,dt,a.dt) in (1,-1))
order by 1
10. 如何写连续三个月有记录的sql
SQL> select org_id , person_id
2 from
3 (
4 select org_id , person_id ,
5 count(*) over ( partition by org_id , person_id order by month_num range 2 preceding ) ttt
6 from test)
7 /
select org_id , person_id
*
ERROR at line 1:
ORA-03113: end-of-file on communication channel