微软的SQL脚本生成令我伤透了心——我一直以为是我的程序上的问题，或者我操作上的失误，并且，客服人员屡屡埋怨我的程序bug，多次测试之后，才发现，原来都是微软惹的祸……

Sql Server的脚本生成有不少漏洞，经常由_art_center>它生成的脚本运行起来会有错误。下面举例说明：

1. 并没有根据sysdenpends的依赖关系生成SQL代码，而是根据所谓的“优先级”来生成。

比如：他认为view的优先级就要比function高。

那么，我写了下面的测试程序，形成如下的依赖关系：fnT1 <-- vwT1 <-- fnT2

就是，view vwT1处于依赖的中间。





_NOBR> _CODE>create function fnT1()
returns Integer
as
begin
return 123
end
go

create view vwT1
as
select aa=dbo.fnT1()

go

create function fnT2()
returns table
as
return (select * from vwT1)
go_CODE>

_NOBR>

运行到数据库之后，用Enterprise生成SQL代码。要注意选项不一样，生成的代码会有所不同，在这里我没有选数据库和用户。

_NOBR> _CODE>if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[fnT1]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[fnT1]
GO

if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[fnT2]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[fnT2]
GO

if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[vwT1]') and OBJECTPROPERTY(id, N'IsView') = 1)
drop view [dbo].[vwT1]
GO

SET QUOTED_IDENTIFIER ON
GO
SET ANSI_NULLS ON
GO

Create view vwT1
As
Select aa=dbo.fnT1()

GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS ON
GO

SET QUOTED_IDENTIFIER ON
GO
SET ANSI_NULLS ON
GO

Create function fnT1()
Returns Integer
As
begin
Return 123
end

GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS ON
GO

SET QUOTED_IDENTIFIER ON
GO
SET ANSI_NULLS ON
GO

Create function fnT2()
Returns Table
As
Return (Select * From vwT1)

GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS ON
GO_CODE>

_NOBR>