SQL Server 2000은 루프 내에서 1024개의 테이블을 생성합니다

관리자 · 게시됨 2015. 2. 9. 오후 2:04:26

declare @i int
declare @j int
declare @sql varchar(4000)
declare @sqlsource_delete varchar(4000)
declare @sqlsource_create varchar(4000)
declare @sn varchar(10)
set @sql = ''
set @i=1
set @sqlsource_delete = 'drop table [dbo].[Test_{SN}]'
set @sqlsource_create = 'CREATE TABLE [dbo].[Test_{SN}] (
[iid] [int] IDENTITY (1, 1) NOT NULL ,
[headid] [int] NULL ,
[istid] [nvarchar] (255) COLLATE Chinese_PRC_CI_AS NULL ,
[istname] [nvarchar] (255) COLLATE Chinese_PRC_CI_AS NULL ,
[isturl] [nvarchar] (255) COLLATE Chinese_PRC_CI_AS NULL ,
[ititle] [nvarchar] (255) COLLATE Chinese_PRC_CI_AS NULL ,
[icontents] [nvarchar] (255) COLLATE Chinese_PRC_CI_AS NULL ,
[icontent] [ntext] COLLATE Chinese_PRC_CI_AS NULL ,
[iurl] [ntext] COLLATE Chinese_PRC_CI_AS NULL ,
[ipic] [ntext] COLLATE Chinese_PRC_CI_AS NULL ,
[ipdate] [nvarchar] (50) COLLATE Chinese_PRC_CI_AS NULL ,
[iedate] [nvarchar] (50) COLLATE Chinese_PRC_CI_AS NULL ,
[icompname] [nvarchar] (50) COLLATE Chinese_PRC_CI_AS NULL ,
[icompemail] [nvarchar] (255) COLLATE Chinese_PRC_CI_AS NULL ,
[iphone] [nvarchar] (255) COLLATE Chinese_PRC_CI_AS NULL ,
[ifax] [nvarchar] (100) COLLATE Chinese_PRC_CI_AS NULL ,
[ishengf] [nvarchar] (50) COLLATE Chinese_PRC_CI_AS NULL ,
[iaddress] [nvarchar] (255) COLLATE Chinese_PRC_CI_AS NULL ,
[ilianxiren] [nvarchar] (50) COLLATE Chinese_PRC_CI_AS NULL ,
[imobile] [nvarchar] (50) COLLATE Chinese_PRC_CI_AS NULL ,
[isite] [nvarchar] (255) COLLATE Chinese_PRC_CI_AS NULL ,
[isamecount] [int] NULL ,
[imainid] [int] NULL ,
[creationDate] [datetime] NOT NULL DEFAULT getdate() ,
[lastModifiedDate] [datetime] NOT NULL DEFAULT getdate()
CONSTRAINT [PK_Test_{SN}] PRIMARY KEY CLUSTERED
(
[iid]
) ON [PRIMARY]
) ON [PRIMARY]'
while @i<=1024
begin
set @sn = cast(@i as varchar(10))
set @j=len(@sn)
if exists (select * from dbo.sysobjects where id = object_id('[dbo].[Test_' + @sn + ']') and OBJECTPROPERTY(id, 'IsUserTable') = 1)
begin
set @sql = replace(@sqlsource_delete, '{SN}', @sn)
--print @sql
execute(@sql)
end
set @sql = replace(@sqlsource_create, '{SN}', @sn)
--print @sql
execute(@sql)
set @i = @i + 1
end

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[출처] SQL Server 2000은 루프 내에서 1024개의 테이블을 생성합니다

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