我有两张桌子
tmpEntityAddress
EntityId Address
________ _______
5 <Address />
5 <Address />
7 <Address />
tmpEntityAddresses
EntityId XML
________ _______
5 <Addresses />
5 <Addresses />
我想将各种地址文档分组并插入第二个表中的单个地址文档。粗略的基本架构看起来像:
<Addresses>
<Address>
<Street />
<PostCode />
</Address>
<Address>
<Street />
<PostCode />
</Address>
</Addresses>
我无法完全理解如何使用XML DML语言功能在SQL中执行此操作,例如XML插入等https://docs.microsoft.com/en-us/sql/t-sql/xml/xml-data-modification-language-xml-dml
我在想我可以做以下事情:
update tmpEntityAddresses
set
XML.modify('insert sql:column("Address") into (/Addresses)[1]'
from
tmpEntityAddresses
join #tmpEntityAddresses on tmpEntityAddresses.EntityId = tmpEntityAddress.EntityId
但似乎只从#tmpEntityAddress添加一行,这不是我想要的,因为我需要整个集合。
这可能与SQL有关吗?如果是这样,怎么能实现呢?
答案 0 :(得分:1)
你是对的,你可以一次只使用.modify()
一个动作......
以下代码有一些假设:
Addresses
中的第二行应该有EntityId=7
您可以使用可更新的CTE 。此CTE将选择Addresses
列,并从Address
表中为给定ID添加所有XML的计算列。
现在我们可以使用modify,一次性在一个动作中插入所有组合的地址条目:
DECLARE @address TABLE (ID INT,[Address] XML);
INSERT INTO @address VALUES
(5, N'<Address id="5a" ><Street>Some Street</Street></Address>')
,(5, N'<Address id="5b" ><Street>Some Other</Street></Address>')
,(7, N'<Address id="7" ><Street>One More</Street></Address>');
DECLARE @addresses TABLE (ID INT,AddrXML XML);
INSERT INTO @addresses VALUES
(5, N'<Addresses><PreexistingContent>Blah</PreexistingContent></Addresses>')
,(7, N'<Addresses><PreexistingContent>Booh</PreexistingContent></Addresses>');
WITH CombinedAddress AS
(
SELECT adrs.ID
,(
SELECT adr.[Address]
FROM @address AS adr
WHERE adr.ID=adrs.ID
FOR XML PATH(''),TYPE
) AS Combined
,adrs.AddrXML
FROM @addresses AS adrs
)
UPDATE CombinedAddress
SET AddrXML.modify(N'insert sql:column("Combined") as last into (/Addresses)[1]');
SELECT * FROM @addresses
ID = 5的结果
<Addresses>
<PreexistingContent>Blah</PreexistingContent>
<Address>
<Address id="5a">
<Street>Some Street</Street>
</Address>
</Address>
<Address>
<Address id="5b">
<Street>Some Other</Street>
</Address>
</Address>
</Addresses>
试试这个:
WITH CombinedAddress AS
(
SELECT adrs.ID
,(
SELECT adr.[Address] AS [*]
FROM @address AS adr
WHERE adr.ID=adrs.ID
FOR XML PATH(''),TYPE
) AS Combined
,adrs.AddrXML
FROM @addresses AS adrs
)
UPDATE CombinedAddress
SET AddrXML.modify(N'insert sql:column("Combined") as last into (/Addresses)[1]');
结果
<Addresses>
<PreexistingContent>Blah</PreexistingContent>
<Address id="5a">
<Street>Some Street</Street>
</Address>
<Address id="5b">
<Street>Some Other</Street>
</Address>
</Addresses>
答案 1 :(得分:0)
你绝对正确,你不应该像字符串一样连接XML。这是通过相关子查询正确完成的方法:
declare @t table (
Id int not null,
Address xml not null
);
insert into @t (Id, Address)
values
(5, N'<Address Val="1" />'),
(5, N'<Address Val="2" />'),
(7, N'<Address Val="7" />');
select sq.Id, (
select t.Address
from @t t
where t.Id = sq.Id
for xml path(''), type, root('Addresses')
)
from (select distinct i.Id from @t i) sq
order by sq.Id;
您可以根据需要在insert
或update
中将结果输出用作来源。