调整编码的隶属关系2,在线等答案

[ynshwsx]

if object_id('hzfa') is not null drop table hzfa
create table hzfa(id int identity,bh varchar(100))
insert hzfa(bh)
select '01'
union all select '0101'
union all select '01017710'
union all select '02'
union all select '0202'
union all select '02020020'
union all select '03'
union all select '0303'
union all select '03030030'



declare
  @tablename sysname,
  @fieldname sysname,
  @CodeRule varchar(50),
  @code varchar(50)  ---要删除的节点编码
select @tablename='hzfa',@fieldname='bh',@CodeRule='1,2,3',@code='01'
 --生成被删除节点的子节点的新编码规则(因为子节点要提升一级)
 declare @codelen int,
         @codelens int,
         @CodeRule_new varchar(50)
 select @codelens=len(@code),
        @CodeRule_new=@CodeRule+','
 --获取被删除的编码的最后一级的长度,并生成该编码之后编码规则
 while @CodeRule_new>'' and @codelens>0
    begin
       select @codelen=LEFT(@CodeRule_new,charindex(',',@CodeRule_new)-1),
              @CodeRule_new=STUFF(@CodeRule_new,1,charindex(',',@CodeRule_new),''),
              @codelens=@codelens-@codelen
    end
        print @codelen          2
        print @CodeRule_new     3,
        print @codelens         -1
       --确认要删除的编码是否符合编码规则
       if @codelens<>0
         begin
             raiserror(N'编码不符合指定的编码规则"%s"',16,1,@code,@CodeRule)
             return
         end   
                   print @CodeRule_new

编码规则1,2,3删除表中01的编码,子编码上升一级,问题是以上代码判断@CODELENS <>0显示错误,执行以上语句之后,@CODELENS不可能为0,请高手指点,在线等

[ai_li7758521]

路过

[needanothercoder]

ding

[-晴天]

把while先注释掉,看只循环一次的结果与你想要的结果是否相同.

[soft_wsx]

楼主错了
比如说编码规则是2,2,3
有如下数据
要删除二级001,应该为01001,不信可以自己看看

if object_id('hzfa') is not null drop table hzfa
create table hzfa(id int identity,bh varchar(100))
insert hzfa(bh)
select '01'
union all select '01001'
union all select '010010001'
union all select '02'
union all select '02002'
union all select '020020002'
union all select '03'
union all select '03003'
union all select '030030003'

[soft_wsx]

楼主可以按照我的方法一步步试试看

declare
  @tablename sysname,
  @fieldname sysname,
  @CodeRule varchar(50),
  @code varchar(50)  ---要删除的节点编码
select @tablename='hzfa',@fieldname='bh',@CodeRule='2,3,4',@code='01001'
 --生成被删除节点的子节点的新编码规则(因为子节点要提升一级)
 declare @codelen int,
         @codelens int,
         @CodeRule_new varchar(50)
 select @codelens=len(@code),
        @CodeRule_new=@CodeRule+','
 --获取被删除的编码的最后一级的长度,并生成该编码之后编码规则
 while @CodeRule_new>'' and @codelens>0
    begin
       select @codelen=LEFT(@CodeRule_new,charindex(',',@CodeRule_new)-1),
              @CodeRule_new=STUFF(@CodeRule_new,1,charindex(',',@CodeRule_new),''),
              @codelens=@codelens-@codelen
    end
        print @codelen
        print @CodeRule_new
        print @codelens
       --确认要删除的编码是否符合编码规则
       if @codelens<>0
         begin
             raiserror(N'编码不符合指定的编码规则"%s"',16,1,@code,@CodeRule)
             return
         end   
       
       --生成修改被删除节点子节点编码的新编码规则
       set @CodeRule_new=case when @CodeRule_new='' then @CodeRule_new
                           else STUFF(@CodeRule,2+len(@CodeRule)-charindex(',',reverse(@CodeRule)+',',len(@CodeRule_new)+2),0,'0,')
                         end
                         print @CodeRule_new
       
       --得到被删除节点子节点编码更新和T-sql
       declare @sql nvarchar(4000)
       set @sql=dbo.f_changecoderule(@coderule,@CodeRule_new,'',-1,@fieldname)
       
       print @sql
       --检查并完成删除处理
       declare @value_code varchar(50),
               @value_code_child varchar(50)
       select @value_code=QUOTENAME(@code,N''''),
              @value_code_child=QUOTENAME(@code+N'_%',N'''')
        print @value_code
        print @value_code_child      

           select no_old=bh, 
                  no_new= case 
                           when len(bh)<1 then ''
                           else substring(bh,1,2)
            end
                       +case 
                           when len(bh)<6 then ''
                           else left(substring(bh,6,4),3)
            end
           into #
             from hzfa with(xlock,tablock)
             where bh like '01001_%'

         delete from hzfa
           where bh='01001'
           
           
           update a set bh=b.no_new
             from hzfa a,# b
             where a.bh=b.no_old
             
             
             
             select * from hzfa
             
             
--结果
1	01
3	01100
4	01200
5	01300
6	02
7	02002
8	020020002
9	03
10	03003
11	030030003

[ynshwsx]

怎么没人回签

[claro]

学习。