xml – 如何在使用带有未知命名空间的XSLT时获取属性值？

我收到第三方提要,我无法确定命名空间,所以我当前不得不在我的XSLT中使用local-name()函数来获取元素值.但是我需要从一个这样的元素中获取一个属性,并且当命名空间未知时我不知道如何执行此操作(因此需要local-name()函数).

注：我使用.net 2.0来处理XSLT

以下是XML的示例：

<?xml version="1.0" encoding="UTF-8"?>
<feed xmlns="http://www.w3.org/2005/Atom">
   <id>some id</id>
   <title>some title</title>
   <updated>2008-09-11T15:53:31+01:00</updated>
   <link rel="self" href="http://www.somefeedurl.co.uk" />
   <author>
      <name>some author</name>
      <uri>http://someuri.co.uk</uri>
   </author>
   <generator uri="http://aardvarkmedia.co.uk/">AardvarkMedia script</generator>
   <entry>
      <id>http://soemaddress.co.uk/branded3/80406</id>
      <title type="html">My Ttile</title>
      <link rel="alternate" href="http://www.someurl.co.uk" />
      <updated>2008-02-13T00:00:00+01:00</updated>
      <published>2002-09-11T14:16:20+01:00</published>
      <category term="mycategorytext" label="restaurant">Test</category>
      <content type="xhtml">
         <div xmlns="http://www.w3.org/1999/xhtml">
            <div class="vcard">
               <p class="fn org">some title</p>
               <p class="adr">
                  <abbr class="type" title="POSTAL" />
                  <span class="street-address">54 Some Street</span>
                  ,
                  <span class="locality" />
                  ,
                  <span class="country-name">UK</span>
               </p>
               <p class="tel">
                  <span class="value">0123456789</span>
               </p>
               <div class="geo">
                  <span class="latitude">51.99999</span>
                  ,
                  <span class="longitude">-0.123456</span>
               </div>
               <p class="note">
                  <span class="type">Review</span>
                  <span class="value">Some content</span>
               </p>
               <p class="note">
                  <span class="type">Overall rating</span>
                  <span class="value">8</span>
               </p>
            </div>
         </div>
      </content>
      <category term="cuisine-54" label="Spanish" />
      <Point xmlns="http://www.w3.org/2003/01/geo/wgs84_pos#">
         <lat>51.123456789</lat>
         <long>-0.11111111</long>
      </Point>
   </entry>
</feed>

这是XSLT

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

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:atom="http://www.w3.org/2005/Atom" xmlns:wgs="http://www.w3.org/2003/01/geo/wgs84_pos#" exclude-result-prefixes="atom wgs">
  <xsl:output method="xml" indent="yes"/>

  <xsl:key name="uniqueVenuesKey" match="entry" use="id"/>
  <xsl:key name="uniqueCategoriesKey" match="entry" use="category/@term"/>

  <xsl:template match="/">
    <locations>
      <!-- Get all unique venues -->
      <xsl:for-each select="/*[local-name()='feed']/*[local-name()='entry']">
        <xsl:variable name="CurrentVenueKey" select="*[local-name()='id']" ></xsl:variable>
        <xsl:variable name="CurrentVenueName" select="*[local-name()='title']" ></xsl:variable>
        <xsl:variable name="CurrentVenueAddress1" select="*[local-name()='content']/*[local-name()='div']/*[local-name()='div']/*[local-name()='p'][@class='adr']/*[local-name()='span'][@class='street-address']" ></xsl:variable>
        <xsl:variable name="CurrentVenueCity" select="*[local-name()='content']/*[local-name()='div']/*[local-name()='div']/*[local-name()='p'][@class='adr']/*[local-name()='span'][@class='locality']" ></xsl:variable>
        <xsl:variable name="CurrentVenuePostcode" select="*[local-name()='postcode']" ></xsl:variable>
        <xsl:variable name="CurrentVenueTelephone" select="*[local-name()='telephone']" ></xsl:variable>
        <xsl:variable name="CurrentVenueLat" select="*[local-name()='Point']/*[local-name()='lat']" ></xsl:variable>
        <xsl:variable name="CurrentVenueLong" select="*[local-name()='Point']/*[local-name()='long']" ></xsl:variable>
        <xsl:variable name="CurrentCategory" select="WHATDOIPUTHERE"></xsl:variable>

            <location>
              <locationName>
                <xsl:value-of select = "$CurrentVenueName" />
              </locationName>
              <category>
                <xsl:value-of select = "$CurrentCategory" />
              </category>
              <description>
                  <xsl:value-of select = "$CurrentVenueName" />
              </description>
              <venueAddress>
                <streetName>
                  <xsl:value-of select = "$CurrentVenueAddress1" />
                </streetName>
                <town>
                  <xsl:value-of select = "$CurrentVenueCity" />
                </town>
                <postcode>
                  <xsl:value-of select = "$CurrentVenuePostcode" />
                </postcode>
                <wgs84_latitude>
                  <xsl:value-of select = "$CurrentVenueLat" />
                </wgs84_latitude>
                <wgs84_longitude>
                  <xsl:value-of select = "$CurrentVenueLong" />
                </wgs84_longitude>
              </venueAddress>
              <venuePhone>
                <phonenumber>
                  <xsl:value-of select = "$CurrentVenueTelephone" />
                </phonenumber>
              </venuePhone>
          </location>
        </xsl:for-each>
    </locations>
  </xsl:template>
</xsl:stylesheet>

我正在尝试将$CurrentCategory变量替换为适当的代码以显示mycategorytext

我这里没有XSLT编辑器,但您尝试过使用

*[local-name()='category']/@*[local-name()='term']