XSLT：在XSLT中创建一个地图 [英] XSLT : Creating a Map in XSLT

问题描述

我想在xsl中有一个键值映射，所以定义了一个具有xml片段的变量，但是后来当我尝试访问变量中的xml节点时，我得到一个错误，类型的xpath不能被解析。

 < xsl：stylesheet version =1.0xmlns：xsl =http://www.w3.org/ 1999 / XSL /变换> 
< xsl：template match =/> 
< xsl：variable name =map> 
< map> 
< entry key =key-1> value1< / entry> 
< entry key =key-2> value2< / entry> 
< entry key =key-3> value3< / entry> 
< / map> 
< / xsl：variable> 
< output> 
< xsl：value-of select =$ map / entry [@ key ='key-1']/> 
< / output> 
< / xsl：template> 
< / xsl：stylesheet> 
  

解决方案

XSLT 2.0

使用XSLT 2.0，以下解决方案的工作原理是：

 < xsl：variable name =map> ; 
< entry key =key-1> value1< / entry> 
< entry key =key-2> value2< / entry> 
< entry key =key-3> value3< / entry> 
< / xsl：variable> 
 
< xsl：template match =/> 
< output> 
< xsl：value-of select =$ map / entry [@ key ='key-1']/> 
< / output> 
< / xsl：template> 
  

XSLT 1.0

您不能使用XSLT 1.0中XPath表达式中的结果树片段，但 fn：document（）可以检索地图值。一个类似的问题将在这里工作：。

 < xsl：value-of select = 文件（ ''）//的xsl：变量[@名称= '映射'] /图/条目[@键= '键-1']/> 
  

如 XSLT 1.0规范：

）是指
的根节点的样式表;样式表的树
表示是
完全相同，如果包含样式表的XML
文档是
的初始源文档。

< blockquote>

但是，您不需要为此使用 xsl：variable 。您可以直接在 xsl：stylesheet 下指定地图节点，但您必须记住，顶级元素必须具有非空名称空间URI：

 < xsl：stylesheet 
 version =1.0xmlns：xsl =http://www.w3.org/1999/XSL/Transform 
 xmlns：my =some.uriexclude-result-prefixes =my> 
 
< my：map> 
< entry key =key-1> value1< / entry> 
< entry key =key-2> value2< / entry> 
< entry key =key-3> value3< / entry> 
< / my：map> 
 
< xsl：template match =/> 
< output> 
< xsl：value-of select =document（''）/ * / my：map / entry [@ key ='key-1']/> 
< / output> 
< / xsl：template> 
< / xsl：stylesheet> 
  

I want to have a key value map in xsl and so defined a variable which has an xml fragment, but later when I try to access the xml nodes in the variable I get an error that type of the xpath xpression cannot be resolved.

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:template match="/">
        <xsl:variable name="map">
            <map>
                <entry key="key-1">value1</entry>
                <entry key="key-2">value2</entry>
                <entry key="key-3">value3</entry>
            </map>
        </xsl:variable>
        <output>
            <xsl:value-of select="$map/entry[@key='key-1']"/>
        </output>
    </xsl:template>
</xsl:stylesheet>

解决方案

XSLT 2.0

Using XSLT 2.0, the following solution works:

  <xsl:variable name="map">
    <entry key="key-1">value1</entry>
    <entry key="key-2">value2</entry>
    <entry key="key-3">value3</entry>
  </xsl:variable>

  <xsl:template match="/">
    <output>
      <xsl:value-of select="$map/entry[@key='key-1']"/>
    </output>
  </xsl:template>

XSLT 1.0

You cannot use a result tree fragment in a XPath expression in XSLT 1.0, but fn:document() can retrieve map values. An answer to a similar question will work here:.

<xsl:value-of select="document('')//xsl:variable[@name='map']/map/entry[@key='key-1']"/>

As described in the XSLT 1.0 specification:

document("") refers to the root node of the stylesheet; the tree representation of the stylesheet is exactly the same as if the XML document containing the stylesheet was the initial source document.

However, you don't need to use xsl:variable for this. You could specify your map node directly under xsl:stylesheet, but you must remember that a top level elements must have a non null namespace URI:

<xsl:stylesheet 
  version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" 
  xmlns:my="some.uri" exclude-result-prefixes="my">

  <my:map>
    <entry key="key-1">value1</entry>
    <entry key="key-2">value2</entry>
    <entry key="key-3">value3</entry>
  </my:map>

  <xsl:template match="/">
    <output>
      <xsl:value-of select="document('')/*/my:map/entry[@key='key-1']"/>
    </output>
  </xsl:template>
</xsl:stylesheet>

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