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ASP 中的 用排序串字段实现树状结构（存储过程）
出处:互联网 整理: 软晨网（RuanChen.com） 发布: 2009-11-03 浏览: 221 ::
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if exists (select * from sysobjects where id = object_id("lybsave"))
drop proc lybsave
CREATE PROCEDURE [lybsave] @keyid int=0,@guestname varchar(20),@guestitle varchar(100),@guestcomm
text,@guestemail varchar(50)='',@emailflag bit=0,@fromip varchar(15),@recimail varchar(50) OUTPUT
AS
DECLARE @ostr varchar(30),@rootid int,@lybid int,@ostrs varchar(30),@l tinyint,@tdt datetime,@putdate
varchar(10),@puttime varchar(5),@eflag bit
select @tdt=getdate()
select @putdate=convert(varchar(4),datepart(yy,@tdt))+'-'+left('0'+convert(varchar(2),datepart(mm,@tdt)),2)
+'-'+left('0'+convert(varchar(2),datepart(dd,@tdt)),2)
select @puttime=left('0'+convert(varchar(2),datepart(hh,@tdt)),2)+':'+left('0'+convert(varchar(2),datepart
(mi,@tdt)),2)
select @ostr='',@rootid=0,@lybid=0,@l=0
if (@guestemail='') select @emailflag=0
If @keyid=0 --发新贴
goto newin
ELSE
begin
SELECT @lybid=lybid,@rootid=rootid,@ostr=orderstr,@recimail=guestemail,@eflag=emailflag from guestbook
where lybid=@keyid
IF @lybid=0 --回复贴没找到，当新贴发表
goto newin
ELSE
BEGIN
if (@eflag=0 and @guestemail<>'[email protected] abc') select @recimail='' --如果是版主回复且指定发邮件给提
问者，则不管发贴者是否要求回复，后面的abc相当于管理密码
if (@rootid=0) select @rootid=@lybid
select @ostrs=@ostr+'%',@lybid=0
select top 1 @lybid=lybid,@ostrs=orderstr from guestbook where rootid=@rootid and (orderstr like
@ostrs) and lybid<>@keyid order by orderstr
if (@lybid=0) select @ostr=@ostr+char(122)
else
begin
select @l=len(@ostrs)
select @ostr=left(@ostrs,@l-1)+char(ascii(substring(@ostrs,@l,1))-1)
end
goto newin
end
end
newin:
INSERT into guestbook
(guestname,guestitle,guestcomm,putdate,puttime,guestemail,emailflag,rootid,fromip,orderstr) values
(@guestname,@guestitle,@guestcomm,@putdate,@puttime,rtrim(@guestemail),@emailflag,@rootid,@fromip,@ostr)
删贴（剪枝）存储过程：
if exists (select * from sysobjects where id = object_id("lybdel"))
drop proc lybdel
CREATE PROCEDURE [lybdel] @keyid int
AS
DECLARE @ostr varchar(30),@rootid int,@lybid int
select @ostr='',@rootid=0,@lybid=0
SELECT @ostr=orderstr,@rootid=rootid,@lybid=lybid from guestbook where lybid=@keyid
if (@lybid<>0)
BEGIN
if (@rootid=0) select @rootid=@lybid
SELECT @ostr=@ostr+'%'
DELETE FROM guestbook where orderstr like @ostr and rootid=@rootid or lybid=@rootid
END
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