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
*/