使用函数返回列表值

时间:2014-03-07 08:23:30

标签: sql-server tsql

我正在尝试编写查询,并且正在使用一个card_nr。例如:

Declare @card_nr as nvarchar(50)

set @card_nr = '704487442952000472'

select 
    a.times, b.times 
from 
    (select 
         stat_id, times 
     from info 
     where card_nr = @card_nr) as a 
left outer join 
    (select 
         stat_id, times 
     from info 
     where card_nr = @card_nr) as b on a.stat_id = b.stat_id+1

我得到了正确的结果:

2014-02-04 11:20:00.000 NULL
2014-02-06 09:44:00.000 2014-02-04 11:20:00.000
2014-02-12 09:59:00.000 2014-02-06 09:44:00.000
2014-02-13 10:31:00.000 2014-02-12 09:59:00.000
2014-02-21 09:49:00.000 2014-02-13 10:31:00.000

我想从表格列表中为每个card_nr执行此操作。所以,首先我写了这个函数:

alter FUNCTION tabletest
   ( @card_nr as nvarchar(50))
RETURNS smalldatetime 
AS BEGIN
    Declare @sqldata as smalldatetime

    select 
        @sqldata = b.times 
    from 
        (select 
             stat_id, times 
         from info where card_nr = @card_nr) as a 
    left outer join 
        (select 
             stat_id, times 
         from info where card_nr = @card_nr ) as b on a.stat_id = b.stat_id + 1

   RETURN @datas 
END

在我尝试的功能之后选择我的查询:

select name, times, dbo.tabletest(card_nr)
from dbo.info

我的结果不正确:

User1   2014-02-04 11:20:00.000 2014-02-21 09:49:00
User1   2014-02-06 09:44:00.000 2014-02-21 09:49:00
User1   2014-02-12 09:59:00.000 2014-02-21 09:49:00
User1   2014-02-13 10:31:00.000 2014-02-21 09:49:00
User1   2014-02-21 09:49:00.000 2014-02-21 09:49:00
User2   2014-02-14 13:41:00.000 2014-02-28 12:20:00
User2   2014-02-24 11:46:00.000 2014-02-28 12:20:00
User2   2014-02-28 12:20:00.000 2014-02-28 12:20:00

我想得到这个:

User1   2014-02-04 11:20:00.000 NULL
User1   2014-02-06 09:44:00.000 2014-02-04 11:20:00.000
User1   2014-02-13 10:31:00.000 2014-02-12 09:59:00.000
User1   2014-02-13 10:31:00.000 2014-02-12 09:59:00.000
User1   2014-02-21 09:49:00.000 2014-02-13 10:31:00.000
User2   2014-02-14 13:41:00.000 NULL
User2   2014-02-24 11:46:00.000 2014-02-14 13:41:00.000
User2   2014-02-28 12:20:00.000 2014-02-24 11:46:00.000

1 个答案:

答案 0 :(得分:0)

应该可以仅使用表格信息获得所需的结果:

 select name, times, 
 (select max(times) from info i2 where i2.times<i1.times and i1.name=i2.name and i2.card_nr=@card_nr ) 
 from info i1
 where  card_nr=@card_nr