SQL code
/*
方法很多，这里简单写一个
返回@find在@str中第(@n)次出现的位置。没有第(@n)次返回0。
*/
create
function
fn_find(@find
varchar
(8000), @str
varchar
(8000), @n
smallint
)
    
returns
int
as
begin
    
if @n < 1
return
(0)
    
declare
@start
smallint
, @
count
smallint
, @
index
smallint
, @len
smallint
    
set
@
index
= charindex(@find, @str)
    
if @
index
= 0
return
(0)
    
else
select
@
count
= 1, @len = len(@find)
    
while @
index
> 0
and
@
count
< @n
        
begin
            
set
@start = @
index
+ @len
            
select
@
index
= charindex(@find, @str, @start), @
count
= @
count
+ 1
        
end
    
if @
count
< @n
set
@
index
= 0
    
return
(@
index
)
end
go
declare
@str
varchar
(100)
set
@str=
'A,B,C,D,A,B,C,D,C,D,B,A,C,E'
select
dbo.fn_find(
'A'
,@str,1)
as
one, dbo.fn_find(
'A'
,@str,2)
as
two, dbo.fn_find(
'A'
,@str,3)
as
three, dbo.fn_find(
'A'
,@str,4)
as
four
/*
one         two         three       four       
----------- ----------- ----------- -----------
1           9           23          0