结果：

Question

更新：我发现针对此问题引发了Microsoft Connect项目here

使用FOR XML PATH和WITH XMLNAMESPACES声明默认命名空间时，我会在使用FOR XML的嵌套查询的任何顶级节点中复制命名空间，我偶然发现了一些解决方案在线，但我并不完全相信......

以下是完整示例

/*
drop table t1
drop table t2
*/
create table t1 ( c1 int, c2 varchar(50))
create table t2 ( c1 int, c2 int, c3 varchar(50))
insert t1 values 
(1, 'Mouse'),
(2, 'Chicken'),
(3, 'Snake');
insert t2 values
(1, 1, 'Front Right'),
(2, 1, 'Front Left'),
(3, 1, 'Back Right'),
(4, 1, 'Back Left'),
(5, 2, 'Right'),
(6, 2, 'Left')



;with XmlNamespaces( default 'uri:animal')
select 
    a.c2 as "@species"
    , (select l.c3 as "text()" 
       from t2 l where l.c2 = a.c1 
       for xml path('leg'), type) as "legs"
from t1 a
for xml path('animal'), root('zoo')

什么是最好的解决方案？

Answer 1

如果我理解正确，您指的是您在查询中可能会看到的行为：

DECLARE @Order TABLE (
  OrderID INT, 
  OrderDate DATETIME)

DECLARE @OrderDetail TABLE (
  OrderID INT, 
  ItemID VARCHAR(1), 
  ItemName VARCHAR(50), 
  Qty INT)

INSERT @Order 
VALUES 
(1, '2010-01-01'),
(2, '2010-01-02')

INSERT @OrderDetail 
VALUES 
(1, 'A', 'Drink',  5),
(1, 'B', 'Cup',    2),
(2, 'A', 'Drink',  2),
(2, 'C', 'Straw',  1),
(2, 'D', 'Napkin', 1)

;WITH XMLNAMESPACES('http://test.com/order' AS od) 
SELECT
  OrderID AS "@OrderID",
  (SELECT 
     ItemID AS "@od:ItemID", 
     ItemName AS "data()" 
   FROM @OrderDetail 
   WHERE OrderID = o.OrderID 
   FOR XML PATH ('od.Item'), TYPE)
FROM @Order o 
FOR XML PATH ('od.Order'), TYPE, ROOT('xml')

其中给出了以下结果：

<xml xmlns:od="http://test.com/order">
  <od.Order OrderID="1">
    <od.Item xmlns:od="http://test.com/order" od:ItemID="A">Drink</od.Item>
    <od.Item xmlns:od="http://test.com/order" od:ItemID="B">Cup</od.Item>
  </od.Order>
  <od.Order OrderID="2">
    <od.Item xmlns:od="http://test.com/order" od:ItemID="A">Drink</od.Item>
    <od.Item xmlns:od="http://test.com/order" od:ItemID="C">Straw</od.Item>
    <od.Item xmlns:od="http://test.com/order" od:ItemID="D">Napkin</od.Item>
  </od.Order>
</xml>

正如您所说，命名空间在子查询的结果中重复。

此行为是根据devnetnewsgroup上的对话（网站现已解散）的功能，尽管vote可以选择更改它。

我建议的解决方案是恢复为FOR XML EXPLICIT：

SELECT
  1 AS Tag,
  NULL AS Parent,
  'http://test.com/order' AS [xml!1!xmlns:od],
  NULL AS [od:Order!2],
  NULL AS [od:Order!2!OrderID],
  NULL AS [od:Item!3],
  NULL AS [od:Item!3!ItemID]
UNION ALL
SELECT 
  2 AS Tag,
  1 AS Parent,
  'http://test.com/order' AS [xml!1!xmlns:od],
  NULL AS [od:Order!2],
  OrderID AS [od:Order!2!OrderID],
  NULL AS [od:Item!3],
  NULL [od:Item!3!ItemID]
FROM @Order 
UNION ALL
SELECT
  3 AS Tag,
  2 AS Parent,
  'http://test.com/order' AS [xml!1!xmlns:od],
  NULL AS [od:Order!2],
  o.OrderID AS [od:Order!2!OrderID],
  d.ItemName AS [od:Item!3],
  d.ItemID AS [od:Item!3!ItemID]
FROM @Order o INNER JOIN @OrderDetail d ON o.OrderID = d.OrderID
ORDER BY [od:Order!2!OrderID], [od:Item!3!ItemID]
FOR XML EXPLICIT

看到这些结果：

<xml xmlns:od="http://test.com/order">
  <od:Order OrderID="1">
    <od:Item ItemID="A">Drink</od:Item>
    <od:Item ItemID="B">Cup</od:Item>
  </od:Order>
  <od:Order OrderID="2">
    <od:Item ItemID="A">Drink</od:Item>
    <od:Item ItemID="C">Straw</od:Item>
    <od:Item ItemID="D">Napkin</od:Item>
  </od:Order>
</xml>

Answer 2

经过几个小时的绝望和数百次试验＆amp;错误，我已经提出了以下解决方案。

我在 root 节点仅上想要只有一个 xmlns属性时遇到了同样的问题。但是我对很多子查询进行了非常困难的查询，仅FOR XML EXPLICIT方法就太麻烦了。所以，是的，我想在子查询中方便FOR XML PATH，并设置我自己的xmlns。

我很好地借用了 8kb＆＃39> 答案的代码，因为它太棒了。为了更好地理解，我稍微调整了一下。这是代码：

DECLARE @Order TABLE (OrderID INT, OrderDate DATETIME)    
DECLARE @OrderDetail TABLE (OrderID INT, ItemID VARCHAR(1), Name VARCHAR(50), Qty INT)    
INSERT @Order VALUES (1, '2010-01-01'), (2, '2010-01-02')    
INSERT @OrderDetail VALUES (1, 'A', 'Drink',  5),
                           (1, 'B', 'Cup',    2),
                           (2, 'A', 'Drink',  2),
                           (2, 'C', 'Straw',  1),
                           (2, 'D', 'Napkin', 1)

-- Your ordinary FOR XML PATH query
DECLARE @xml XML = (SELECT OrderID AS "@OrderID",
                        (SELECT ItemID AS "@ItemID", 
                                Name AS "data()" 
                         FROM @OrderDetail 
                         WHERE OrderID = o.OrderID 
                         FOR XML PATH ('Item'), TYPE)
                    FROM @Order o 
                    FOR XML PATH ('Order'), ROOT('dummyTag'), TYPE)

-- Magic happens here!       
SELECT 1 AS Tag
      ,NULL AS Parent
      ,@xml AS [xml!1!!xmltext]
      ,'http://test.com/order' AS [xml!1!xmlns]
FOR XML EXPLICIT

结果：

<xml xmlns="http://test.com/order">
  <Order OrderID="1">
    <Item ItemID="A">Drink</Item>
    <Item ItemID="B">Cup</Item>
  </Order>
  <Order OrderID="2">
    <Item ItemID="A">Drink</Item>
    <Item ItemID="C">Straw</Item>
    <Item ItemID="D">Napkin</Item>
  </Order>
</xml>

如果您单独选择@xml，则会看到它包含根节点dummyTag。我们不需要它，因此我们在xmltext查询中使用directive FOR XML EXPLICIT将其删除：

,@xml AS [xml!1!!xmltext]

虽然MSDN中的解释听起来更复杂，但实际上它告诉解析器选择XML根节点的内容。

不确定查询的速度有多快，但目前我正在放松地喝苏格兰威士忌，同时安静地看着代码......

Answer 3

我看到的另一种解决方案是在将xml构建为临时变量后添加XMLNAMESPACES声明：

declare @xml as xml;
select @xml = (
select 
    a.c2 as "@species"
    , (select l.c3 as "text()" 
       from t2 l where l.c2 = a.c1 
       for xml path('leg'), type) as "legs"
from t1 a
for xml path('animal'))

;with XmlNamespaces( 'uri:animal' as an)
select @xml for xml path('') , root('zoo');

Answer 4

此处的问题因使用XML PATH时无法手动直接声明命名空间而更加复杂。 SQL Server将禁止以“xmlns”开头的任何属性名称以及包含冒号的任何标记名称。

我不是不得不求助于使用相对不友好的XML EXPLICIT，而是首先使用'隐形'命名空间定义和引用生成XML，然后按如下所示进行字符串替换......

DECLARE @Order TABLE (
  OrderID INT, 
  OrderDate DATETIME)

DECLARE @OrderDetail TABLE (
  OrderID INT, 
  ItemID VARCHAR(1), 
  ItemName VARCHAR(50), 
  Qty INT)

INSERT @Order 
VALUES 
(1, '2010-01-01'),
(2, '2010-01-02')

INSERT @OrderDetail 
VALUES 
(1, 'A', 'Drink',  5),
(1, 'B', 'Cup',    2),
(2, 'A', 'Drink',  2),
(2, 'C', 'Straw',  1),
(2, 'D', 'Napkin', 1)

declare @xml xml

set @xml = (SELECT
  'http://test.com/order' as "@xxmlns..od",  -- 'Cloaked' namespace def
  (SELECT OrderID AS "@OrderID", 
    (SELECT 
      ItemID AS "@od..ItemID", 
      ItemName AS "data()" 
     FROM @OrderDetail 
     WHERE OrderID = o.OrderID 
     FOR XML PATH ('od..Item'), TYPE)
   FROM @Order o
   FOR XML PATH ('od..Order'), TYPE)
  FOR XML PATH('xml'))

set @xml = cast(replace(replace(cast(@xml as nvarchar(max)), 'xxmlns', 'xmlns'),'..',':') as xml)

select @xml

要指出的一些事项：

我使用'xxmlns'作为'xmlns'和'..'的隐形版本代表'：'。如果您可能将'..'作为文本值的一部分，这可能对您不起作用 - 只要您选择一个有效的XML标识符，就可以用其他内容替换它。
由于我们希望顶层的xmlns定义，我们不能使用XML ROH的'ROOT'选项 - 而是我需要为subselect结构添加另一个外层来实现这一目的。

Answer 5

我在解释所有这些解释时有点混乱，同时手动声明“xmlns：animals”正在完成这项工作：这里是我写的一个例子来生成Open graph元数据

DECLARE @l_xml as XML;
SELECT @l_xml = 
(
SELECT 'http://ogp.me/ns# fb: http://ogp.me/ns/fb# scanilike: http://ogp.me/ns/fb/scanilike#' as 'xmlns:og',
    (SELECT
        (SELECT 'og:title' as 'property', title as 'content' for xml raw('meta'), TYPE),
        (SELECT 'og:type' as 'property', OpenGraphWebMetadataTypes.name as 'content' for xml raw('meta'), TYPE),
        (SELECT 'og:image' as 'property', image as 'content' for xml raw('meta'), TYPE),
        (SELECT 'og:url' as 'property', url as 'content' for xml raw('meta'), TYPE),
        (SELECT 'og:description' as 'property', description as 'content' for xml raw('meta'), TYPE),
        (SELECT 'og:site_name' as 'property', siteName as 'content' for xml raw('meta'), TYPE),
        (SELECT 'og:appId' as 'property', appId as 'content' for xml raw('meta'), TYPE)
     FROM OpenGraphWebMetaDatas INNER JOIN OpenGraphWebMetadataTypes ON OpenGraphWebMetaDatas.type = OpenGraphWebMetadataTypes.id WHERE THING_KEY = @p_index 
     for xml path('header'), TYPE),
     (SELECT '' as 'body' for xml path(''), TYPE)
     for xml raw('html'), TYPE
)

RETURN @l_xml

返回预期结果

<html xmlns:og="http://ogp.me/ns# fb: http://ogp.me/ns/fb# scanilike: http://ogp.me/ns/fb/scanilike#">
<header>
<meta property="og:title" content="The First object"/>
<meta property="og:type" content="scanilike:tag"/>
<meta property="og:image" content="http://www.mygeolive.com/images/facebook/facebook-logo.jpg"/>
<meta property="og:url" content="http://www.scanilike.com/opengraph?id=1"/>
<meta property="og:description" content="This is the very first object created using the IOThing &amp; ScanILike software. We keep it in file for history purpose. "/>
<meta property="og:site_name" content="http://www.scanilike.com"/>
<meta property="og:appId" content="200270673369521"/>
</header>
<body/>
</html>

希望这有助于人们在网上搜索类似的问题。 ; - ）

Answer 6

如果FOR XML PATH实际上更干净地工作，那就太好了。使用@table变量重做原始示例：

declare @t1 table (c1 int, c2 varchar(50));
declare @t2 table (c1 int, c2 int, c3 varchar(50));
insert @t1 values 
    (1, 'Mouse'),
    (2, 'Chicken'),
    (3, 'Snake');
insert @t2 values
    (1, 1, 'Front Right'),
    (2, 1, 'Front Left'),
    (3, 1, 'Back Right'),
    (4, 1, 'Back Left'),
    (5, 2, 'Right'),
    (6, 2, 'Left');

;with xmlnamespaces( default 'uri:animal')
select  a.c2 as "@species",
    (
        select  l.c3 as "text()"
        from    @t2 l
        where   l.c2 = a.c1
        for xml path('leg'), type
    ) as "legs"
from @t1 a
for xml path('animal'), root('zoo');

使用重复的名称空间声明来解决问题XML：

<zoo xmlns="uri:animal">
  <animal species="Mouse">
    <legs>
      <leg xmlns="uri:animal">Front Right</leg>
      <leg xmlns="uri:animal">Front Left</leg>
      <leg xmlns="uri:animal">Back Right</leg>
      <leg xmlns="uri:animal">Back Left</leg>
    </legs>
  </animal>
  <animal species="Chicken">
    <legs>
      <leg xmlns="uri:animal">Right</leg>
      <leg xmlns="uri:animal">Left</leg>
    </legs>
  </animal>
  <animal species="Snake" />
</zoo>

您可以使用具有通配符名称空间匹配（即*：elementName）的XQuery在名称空间之间迁移元素，如下所示，但是对于复杂的XML来说可能非常麻烦：

;with xmlnamespaces( default 'http://tempuri.org/this/namespace/is/meaningless' )
select (
    select  a.c2 as "@species",
        (
            select  l.c3 as "text()"
            from    @t2 l
            where   l.c2 = a.c1
            for xml path('leg'), type
        ) as "legs"
    from @t1 a
    for xml path('animal'), root('zoo'), type
).query('declare default element namespace "uri:animal";
<zoo>
{ for $a in *:zoo/*:animal return
    <animal>
    {attribute species {$a/@species}}
    { for $l in $a/*:legs return
        <legs>
        { for $m in $l/*:leg return
            <leg>{ $m/text() }</leg>
        }</legs>
    }</animal>
}</zoo>');

哪个会产生您想要的结果：

<zoo xmlns="uri:animal">
  <animal species="Mouse">
    <legs>
      <leg>Front Right</leg>
      <leg>Front Left</leg>
      <leg>Back Right</leg>
      <leg>Back Left</leg>
    </legs>
  </animal>
  <animal species="Chicken">
    <legs>
      <leg>Right</leg>
      <leg>Left</leg>
    </legs>
  </animal>
  <animal species="Snake" />
</zoo>

使用FOR XML PATH时，如何在嵌套查询中删除冗余命名空间

6 个答案:

结果：