--排列位置的sql server函数-------------
SET QUOTED_IDENTIFIER ON
GO
SET ANSI_NULLS ON
GO
create procedure up_get_sort
@ls_o varchar(10)
as
declare @ll_sort int,
@ldc_value dec(9,2),
@ls_id varchar(16),
@ll_same int,
@ls_sql varchar(200)
Begin
---排位计算-------
-----------------------
---有这么一个规则 数值相同的排名相同，
---而且占有位置例如第1名分数相同有3人,那么第2名就从4名开始------
--======================---------------
--开始之前要把关键字和数值保存到exam_use_sort表里
set @ll_sort =1
if lower(@ls_o) = @#desc@# or lower(@ls_o)=@#d@#
begin--降序
set @ls_sql = @# SELECT id,use_value FROM exam_use_sort order by isnull(use_value,0) desc@#
end
else
begin--升序
set @ls_sql = @# SELECT id,use_value FROM exam_use_sort order by isnull(use_value,0)@#
end
exec(@#declare cur cursor for @#+@ls_sql)
OPEN cur ;
fetch next from cur into @ls_id,@ldc_value;
WHILE @@FETCH_STATUS = 0
BEGIN
update exam_use_sort set use_sort = @ll_sort where
use_value = @ldc_value;
--有多少条相同的
select @ll_same = isnull(count(*),0) from exam_use_sort
where use_value = @ldc_value;
set @ll_sort = @ll_sort + @ll_same
fetch next from cur into @ls_id,@ldc_value;
END
CLOSE cur;
DEALLOCATE cur;
End
GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS ON
GO
-------函数用到得表------------------
CREATE TABLE [exam_use_sort] (
[id] [varchar] (20) COLLATE Chinese_PRC_CI_AS NOT NULL ,
[use_value] [decimal](18, 2) NULL ,
[use_sort] [int] NULL
) ON [PRIMARY]
GO
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