Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
923 views
in Technique[技术] by (71.8m points)

xslt - XSL - How to remove unused namespaces from source xml?

I have an xml with a lot of unused namespaces, like this:

<?xml version="1.0" encoding="UTF-8"?>
<ns1:Envelope xmlns:ns1="http://www.a.com" xmlns:ns2="http://www.b.com" xmlns:ns3="http://www.c.com" xmlns:ns4="http://www.d.com">
    <ns1:Body>
        <ns2:a>
            <ns2:b>data1</ns2:b>
            <ns2:c>data2</ns2:c>
        </ns2:a>
    </ns1:Body>
</ns1:Envelope> 

I would like to remove the unused namespaces without having to specify in the xslt which ones to remove/maintain. The result xml should be this:

<?xml version="1.0" encoding="UTF-8"?>
<ns1:Envelope xmlns:ns1="http://www.a.com" xmlns:ns2="http://www.b.com">
    <ns1:Body>
        <ns2:a>
            <ns2:b>data1</ns2:b>
            <ns2:c>data2</ns2:c>
        </ns2:a>
    </ns1:Body>
</ns1:Envelope> 

I've googled a lot but haven't found a solution to this particular issue. Is there any?

Thanks.

PS: Not 100% sure but I think it should be for XSL 1.0.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Unlike the answer of @Martin-Honnen, this solution produces exactly the desired result -- the necessary namespace nodes remain where they are and are not moved down.

Also, this solution correctly deals with attributes that are in a namespace:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

 <xsl:template match="node()|@*" priority="-2">
     <xsl:copy>
       <xsl:apply-templates select="node()|@*"/>
     </xsl:copy>
 </xsl:template>

 <xsl:template match="*">
  <xsl:element name="{name()}" namespace="{namespace-uri()}">
   <xsl:variable name="vtheElem" select="."/>

   <xsl:for-each select="namespace::*">
     <xsl:variable name="vPrefix" select="name()"/>

     <xsl:if test=
      "$vtheElem/descendant::*
              [(namespace-uri()=current()
             and 
              substring-before(name(),':') = $vPrefix)
             or
              @*[substring-before(name(),':') = $vPrefix]
              ]
      ">
      <xsl:copy-of select="."/>
     </xsl:if>
   </xsl:for-each>
   <xsl:apply-templates select="node()|@*"/>
  </xsl:element>
 </xsl:template>
</xsl:stylesheet>

when this transformation is applied on the following XML document (the provided XML document with an added namespaced attribute):

<ns1:Envelope xmlns:ns1="http://www.a.com" xmlns:ns2="http://www.b.com" xmlns:ns3="http://www.c.com" xmlns:ns4="http://www.d.com">
    <ns1:Body ns2:x="1">
        <ns2:a>
            <ns2:b>data1</ns2:b>
            <ns2:c>data2</ns2:c>
        </ns2:a>
    </ns1:Body>
</ns1:Envelope>

the desired, correct result is produced:

<ns1:Envelope xmlns:ns1="http://www.a.com" xmlns:ns2="http://www.b.com">
   <ns1:Body ns2:x="1">
      <ns2:a>
         <ns2:b>data1</ns2:b>
         <ns2:c>data2</ns2:c>
      </ns2:a>
   </ns1:Body>
</ns1:Envelope>

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...