den*_*ily 3 sql sql-server select
我有一个包含以下示例数据的表:
Tag Loc Time1
A 10 6/2/15 8:00 AM
A 10 6/2/15 7:50 AM
A 10 6/2/15 7:30 AM
A 20 6/2/15 7:20 AM
A 20 6/2/15 7:15 AM
B 10 6/2/15 7:12 AM
B 10 6/2/15 7:11 AM
A 10 6/2/15 7:10 AM
A 10 6/2/15 7:00 AM
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我需要SQL来选择序列中第一个(最早的)行,直到位置发生变化,然后再次选择最早的行,直到位置发生变化.换句话说,我需要以上输出:
Tag Loc Time1
A 10 6/2/15 7:30 AM
A 20 6/2/15 7:15 AM
A 10 6/2/15 7:00 AM
B 10 6/2/15 7:11 AM
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我在Giorgos尝试了这个 - 但是选择中的一些行是重复的:
declare @temptbl table (rowid int primary key identity, tag nvarchar(1), loc int, time1 datetime)
declare @tag as nvarchar(1), @loc as int, @time1 as datetime
insert into @temptbl (tag, loc, time1) values (1,20,'6/5/2015 7:15 AM')
insert into @temptbl (tag, loc, time1) values (1,20,'6/5/2015 7:20 AM')
insert into @temptbl (tag, loc, time1) values (1,20,'6/5/2015 7:25 AM')
insert into @temptbl (tag, loc, time1) values (4,20,'6/5/2015 7:20 AM')
insert into @temptbl (tag, loc, time1) values (4,20,'6/5/2015 7:25 AM')
insert into @temptbl (tag, loc, time1) values (4,20,'6/5/2015 7:30 AM')
insert into @temptbl (tag, loc, time1) values (4,20,'6/5/2015 7:35 AM')
insert into @temptbl (tag, loc, time1) values (4,20,'6/5/2015 7:40 AM')
select * from @temptbl
SELECT Tag, Loc, MIN(Time1) as time2
FROM (
SELECT Tag, Loc, Time1,
ROW_NUMBER() OVER (ORDER BY Time1) -
ROW_NUMBER() OVER (PARTITION BY Tag, Loc
ORDER BY Time1) AS grp
FROM @temptbl ) t
GROUP BY Tag, Loc, grp
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结果如下(每个标签应该只有一行)
Tag Loc time2
1 20 2015-06-05 07:15:00.000
1 20 2015-06-05 07:25:00.000
4 20 2015-06-05 07:20:00.000
4 20 2015-06-05 07:30:00.000
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假设您正在使用MS SQL Server 2012或更高版本,lag窗口函数将允许您将行与前一行进行比较:
SELECT tag, loc, time1
FROM (SELECT tag, loc, time1,
LAG (loc) OVER (PARTITION BY tag ORDER BY time1) AS lagloc
FROM my_table) t
WHERE loc != lagloc OR lagloc IS NULL
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