Question

我有一个包含4条记录的表格AddressTypes（家庭，办公室，度假，酒店）和一个共享提交的addresstypeid的表格地址。

在地址表中，我有1条“home”类型的记录，我想要一个查询，我得到4行这样：

Type     Address1  Address2 City    State

home     piping    1232     Austin  Tx
office   null      null     null    null
vacation null      null     null    null
hotel    null      null     null    null

以下是表格的图像：http://tinypic.com/view.php?pic=28078xv&s=6

我确信使用交叉连接可能非常容易，但是没有得到它。希望有人能指导我。提前欣赏。

Answer 1

Left joining AdddressTypes到地址将产生所需的结果：

select at.Type, 
       a.Address1, 
       a.Address2, 
       a.City, 
       a.State
  from AddressTypes at
  left join Address a
    on at.AddressTypeID = a.AddressTypeID
-- For this query to make sense
-- Filter one person only
   and a.PersonID = @PersonID

----------- VAAA已添加此部分------------------------

尼古拉，我改变了这个：

select at.description, 
       a.Address1, 
       a.Address2, 
       a.City
  from address_types at
  left join Address a
    on 1 = 1
-- For this query to make sense
-- Filter one person only
   and a.addressid = 24

然后我获得了4行，但是所有这些行都有相同的地址信息，而“home”类型地址就是带有数据的地址。所以它的关闭......

Answer 2

使用LEFT OUTER JOIN，它返回左右表中匹配记录的一个副本和两个表中的不匹配记录。这是我试过的测试代码：

CREATE TABLE AddressTypes(
    AddressType SMALLINT NOT NULL,
    [Description] NVARCHAR(50))

ALTER TABLE AddressTypes
ADD CONSTRAINT PK_AddressTypes_AddressType
    PRIMARY KEY (AddressType)


CREATE TABLE [Address] (
    AddressID INT NOT NULL,
    AddressType SMALLINT,
    Address1 NVARCHAR(50),
    Address2 NVARCHAR(50),
    City NVARCHAR(50),
    State CHAR(2))

ALTER TABLE [Address]
ADD CONSTRAINT PK_Address_AddressID
    PRIMARY KEY (AddressID)

ALTER TABLE [Address]
ADD CONSTRAINT FK_address_addresstypes_addresstype 
    FOREIGN KEY (AddressType) REFERENCES AddressTypes(AddressType)


INSERT INTO AddressTypes VALUES (1, 'home'), (2, 'office'), 
                                (3, 'vacation'), (4, 'hotel')
INSERT INTO address VALUES (1, 1, 'piping', '1232', 'Austin', 'TX')

   -- Here is the query that outputs the result set.
    SELECT AT.AddressType AS [Type], A.Address1, A.Address2, A.City, A.State
    FROM AddressTypes AT
    LEFT OUTER JOIN address A
    ON AT.AddressType = A.AddressType

SQL Cross使用空值连接

2 个答案: