Question

我正在尝试创建一个SQL视图，它整合了许多单独的选择查询。我在将单个select语句中的子句放入数据库视图时遇到了一些困难。

我的观点的简化版本是：

create or replace view TestView as
select 
 A.Name,
 B.Subscription,
 C.Expiry
from
 TestTableA as A left outer join TestTableB as B on A.ID = B.A_ID
 left outer join TestTableC as C on A.ID = C.A_ID;

我对视图有两个问题：

在frist join上我怎样才能选择Subscription是特定值的记录，如果不是那个值仍然检索Name和Expiry列（在这种情况下，Subscription将为null）？
在第二次加入时，如何指定我只想要具有最近到期日期的记录？

以下是我的测试架构，示例数据和所需的结果集：

create table TestTableA
(
    ID int,
    Name varchar(32),
    Primary Key(ID)
);

create table TestTableB
(
    ID int,
    A_ID int,
    Subscription varchar(32),
    Primary Key(ID),
    Foreign Key(A_ID) references TestTableA(ID)
);

create table TestTableC
(
    ID int,
    A_ID int,
    Expiry date,
    Primary Key(ID),
    Foreign Key(A_ID) references TestTableA(ID)
);

create or replace view TestView as
select 
 A.Name,
 B.Subscription,
 C.Expiry
from
 TestTableA as A left outer join TestTableB as B on A.ID = B.A_ID
 left outer join TestTableC as C on A.ID = C.A_ID;

insert into TestTableA values (1, 'Joe');
insert into TestTableB values (1, 1, 'abcd');
insert into TestTableB values (2, 1, 'efgh');
insert into TestTableC values (1, 1, '2012-10-25');
insert into TestTableC values (2, 1, '2012-10-24');
insert into TestTableA values (2, 'Jane');

期望的结果1：

select * from TestView where Subscription is null or Subscription = 'efgh';

Joe, efgh, 2012-10-25
Jane, ,

期望的结果2：

select * from TestView where Subscription is null or Subscription = 'xxxx';

Joe, , 2012-10-25
Jane, ,

Answer 1

我会用简单的SQL写查询如果您有SQL Server 2005或更高版本，则可以使用outer apply而不是使用min()

加入子查询

select 
    A.Name,
    B.Subscription,
    C.Expiry
from TestTableA as A
    left outer join TestTableB as B on A.ID = B.A_ID and B.Subscription in ('abcd', 'efgh') 
    left outer join
    (
        select min(T.Expiry) as Expiry, T.A_ID
        from TestTableC as T
        group by T.A_ID
    ) as C on A.ID = C.A_ID

Answer 2

create or replace view TestView as
select 
  A.Name,
  B.Subscription,
  C.Expiry
from
  TestTableA as A left outer join TestTableB as B on A.ID = B.A_ID
  left outer join TestTableC as C on A.ID = C.A_ID;
where 
  B.Subscription is not null
  and C.Expiry between (now() - interval 1 minute) and now()

可以在视图中使用连接上的where子句

2 个答案: