declare @c varchar(50)set @c='572a3d51-ef7a-459e-a5cd-ebf0fca51e8b' --能查出来呀 你试试,我试一下,好像可以啦谢谢 declare @a nvarchar(50) select @a=Name from DeviceName where PerssionID=1 declare @b int ;with kk AS( select * from DeviceName where ID =@c union all select…