具有XML命名空间的Xsl [英] Xsl with XML namespace

问题描述

当xml具有命名空间时，我如何在xml节点上选择。



这里没有命名空间的简单xml。

How i select on xml node when xml have namespace.

Here simple xml without namespace.

<?xml version="1.0" encoding="utf-8"?>
<HouseList xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <Name>Test</Name>
  <Id>42</Id>
  <HList>
    <House>
      <Name>Home</Name>
      <Id>88</Id>
      <RoomList>
        <Room>
          <Name>Kitchen</Name>
          <Id>21</Id>
        </Room>
      </RoomList>
    </House>
  </HList>
</HouseList>

我的xsl是：

my xsl is:

<pre lang="xml">
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:template match="/">
 <Root>
	<List>
		<HouseListName> 
			<Name>
				<xsl:value-of select="HouseList/Name"/>
			</Name>
			<Rooms>
				<xsl:for-each select="HouseList/HList">
					<Ho>				
						<Name>
							<xsl:value-of select="House/Name"/>
						</Name>
						<R>
							<xsl:for-each select="House/RoomList">
								<Room>				
									<Name>
										<xsl:value-of select="Room/Name"/>	
									</Name>
								</Room>	
							</xsl:for-each>
						</R>					
					</Ho>					
				</xsl:for-each>			
			</Rooms>
		</HouseListName>
	</List>
</Root>
</xsl:template>
</xsl:stylesheet> 

C＃转换

Tranformation with C#

XslCompiledTransform xslt = new XslCompiledTransform();
xslt.Load("xslTest.xsl");
xslt.Transform("Ser.xml", "Result.txt");

我的结果是：

my result is:

<?xml version="1.0" encoding="utf-8"?><Root><List><HouseListName><Name>Test</Name><Rooms><Ho><Name>Home</Name><R><Room><Name>Kitchen</Name></Room></R></Ho></Rooms></HouseListName></List></Root>

一切顺利且有效。

现在我序列化datacontract我的班级因为我现在有一个类的字典。



我的xml现在是：

all good and works.
Now i serialize datacontract my class cause i have now a dictionary in one of a class.

My xml is now:

<?xml version="1.0" encoding="utf-8"?><HouseList xmlns="http://schemas.datacontract.org/2004/07/Xsl_Transformer.Data" xmlns:i="http://www.w3.org/2001/XMLSchema-instance"><HList><House><Id>88</Id><Name>Home</Name><Note xmlns:a="http://schemas.microsoft.com/2003/10/Serialization/Arrays"><a:KeyValueOfstringstring><a:Key>Sinn</a:Key><a:Value>42</a:Value></a:KeyValueOfstringstring><a:KeyValueOfstringstring><a:Key>Egal</a:Key><a:Value>88</a:Value></a:KeyValueOfstringstring></Note><RoomList><Room><Id>21</Id><Name>Kitchen</Name></Room></RoomList></House></HList><Id>42</Id><Name>Test</Name></HouseList>

如何修改我的xsl以进行正确的转换？

如果键Sinn我如何获得价值？



我尝试了什么：



直接使用xsl - xml的唯一节点：

How i must modify my xsl for correct transformation?
How i get access to value if key "Sinn"?

What I have tried:

direct using of xsl - only node of xml:

<?xml version="1.0" encoding="utf-8"?><Root><List><HouseListName><Name></Name><Rooms /></HouseListName></List></Root>

我试试以下网站：

XSLT转换XML w命名空间 - 堆栈溢出 [ ^ ]



i试试这个

I have try following site:
XSLT Transform XML with Namespaces - Stack Overflow[^]

i try this

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"

xmlns:d="http://schemas.datacontract.org/2004/07/Xsl_Transformer.Data"

xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:template match="/d">
 <Root>
	<List>
		<HouseListName> 
			<Name>
				<xsl:value-of select="d:HouseList/Name"/>
			</Name>
			<Rooms>
				<xsl:for-each select="d:HouseList/HList">
					<Ho>				
						<Name>
							<xsl:value-of select="d:House/Name"/>
						</Name>
						<R>
							<xsl:for-each select="d:House/RoomList">
								<Room>				
									<Name>
										<xsl:value-of select="d:Room/Name"/>	
									</Name>
								</Room>	
							</xsl:for-each>
						</R>					
					</Ho>					
				</xsl:for-each>			
			</Rooms>
		</HouseListName>
	</List>
</Root>
</xsl:template>
</xsl:stylesheet> 

i得到这个

i get this

<?xml version="1.0" encoding="utf-8"?>88HomeSinn42Egal8821Kitchen42Test

而没有d匹配 - 我的意思是匹配=/与d命名空间

i得到这个

and without d on match - i mean match="/" with d namespace
i get this

<?xml version="1.0" encoding="utf-8"?><Root xmlns:d="http://schemas.datacontract.org/2004/07/Xsl_Transformer.Data"><List><HouseListName><Name></Name><Rooms /></HouseListName></List></Root>

i读这个，但这甚至不起作用

xslt - 如何从带名称空间的XML中选择？ - 堆栈溢出 [ ^ ]



我能做什么？有人为我链接或帮助吗？

i read this, but this even not work
xslt - How to 'select' from XML with namespaces? - Stack Overflow[^]

What i can do? Have someone link or help for me?

推荐答案

我有解决方案。 我有解决方案 [ ^ ]很好。你必须在每个节点上添加前缀。



喜欢这个：

I have solution. The tip of I have solution[^] was good. You must on every node add prefix.

Like in this:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"

xmlns:d="http://schemas.datacontract.org/2004/07/Xsl_Transformer.Data"

xmlns:xsl="http://www.w3.org/1999/XSL/Transform" exclude-result-prefixes="d">
 
<xsl:template match="/d:HouseList">
 <Root>
	<List>
		<HouseListName> 
			<Name>
				<xsl:value-of select="d:Name"/>
			</Name>
			<Rooms>
				<xsl:for-each select="d:HList">
					<Ho>				
						<Name>
							<xsl:value-of select="d:House/d:Name"/>
						</Name>
						<R>
							<xsl:for-each select="d:House/d:RoomList">
								<Room>				
									<Name>
										<xsl:value-of select="d:Room/d:Name"/>	
									</Name>
								</Room>	
							</xsl:for-each>
						</R>					
					</Ho>					
				</xsl:for-each>			
			</Rooms>
		</HouseListName>
	</List>
</Root>
</xsl:template>
</xsl:stylesheet> 

Result is:

Result is:

<?xml version="1.0" encoding="UTF-8"?><Root><List><HouseListName><Name>Test</Name><Rooms><Ho><Name>Home</Name><R><Room><Name>Kitchen</Name></Room></R></Ho></Rooms></HouseListName></List></Root>

Solved! =)



exclude-result-prefixes remove namespace from the xsl in the new xml output file



How I make it the question as solved?

Solved! =)

exclude-result-prefixes remove namespace from the xsl in the new xml output file

How I make it the question as solved?

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