XML namespace handling in xslt

Question

i need to convert from one xml format to another xml using xslt.Initially i tried without namespace and i got it. but when i tried with name space xmlns="http://ws.wso2.org/dataservice" it not working without prefixes as xmlns:d="http://ws.wso2.org/dataservice" in the given xml file

<test xmlns="http://ws.wso2.org/dataservice">
 <datarows>
  <name>Name</name>
 </datarows>
 <datarows>
  <name>karthik</name>
 </datarows>
 </testcsv>

i need xml file after xsl transformation as like

<head>
<names>Name</names>
<names>karthik</names>
</head>

Help me with xslt

i tried xslt as

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:d="http://ws.wso2.org/dataservice" exclude-result-prefixes="d" >
    <xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>
     <namespace-alias stylesheet-prefix="xsl" result-prefix="#default"/>


    <xsl:template match="/">
    <head>
        <xsl:for-each select="testcsv/datarows">
            <names>
                <xsl:value-of select="name" />
            </names>
        </xsl:for-each>
        </head>
    </xsl:template>
</xsl:stylesheet>

Did you try searching for previous questions on this topic? This question is asked nearly every day, and the answers are easy to find. — Michael Kay
– Michael Kay, Commented Apr 18, 2013 at 10:43

siva2012 · Accepted Answer · 2013-04-18 05:51:26Z

Please change the below list of things: 1. Your Input XML is not well formed. Please Change the root element to <testcsv> 2. Remove <namespace-alias stylesheet-prefix="xsl" result-prefix="#default"/> from your XSLT as it is not needed. 3. Add namespace xmlns:d="http://ws.wso2.org/dataservice" in your Input XML

Input XML:

<?xml version="1.0"?>
<testcsv xmlns:d="http://ws.wso2.org/dataservice">
 <datarows>
  <name>Name</name>
 </datarows>
 <datarows>
  <name>karthik</name>
 </datarows>
 </testcsv>

XSLT transformation:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:d="http://ws.wso2.org/dataservice" exclude-result-prefixes="d" >
    <xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>

    <xsl:template match="/">
    <head>
        <xsl:for-each select="testcsv/datarows">
            <names>
                <xsl:value-of select="name" />
            </names>
        </xsl:for-each>
        </head>
    </xsl:template>
</xsl:stylesheet>

Output:

<head>
  <names>Name</names>
  <names>karthik</names>
</head>

hr_117 · Accepted Answer · 2013-04-18 08:59:57Z

As I understand you would (or could) not change the xmlns declaration in your xml (perhaps because it is generated somewhere else). Than you have to use the xml node prefix in your xlst.

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:d="http://ws.wso2.org/dataservice"
    exclude-result-prefixes="d"
    >
    <xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>
    <xsl:template match="/">
        <head>
            <xsl:for-each select="d:testcsv/d:datarows" >
                <names>
                    <xsl:value-of select="d:name" />
                </names>
            </xsl:for-each>
        </head>
    </xsl:template>
</xsl:stylesheet>

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