Question

i have some requirement, where i need to replace the references in the same xml file.
limitation is to use xslt 1.0 only.

below is my sample input xml.

    <org>
       <depts>
          <dept>
             <deptId>1009</deptId>
                <deptName>IT</deptName>
                <deptAccessCode>IT-1009</deptAccessCode>
          </dept>
          <dept>
             <deptId>2344</deptId>
             <deptName>BPO</deptName>
             <deptAccessCode>BP-2344</deptAccessCode>
          </dept>
          </depts>
          <employees>
             <employee>
             <name>abc</name>
             <dept>
              <REFERENCE>
                 <LocationXPath>/org/depts/dept[2]</LocationXPath>
              </REFERENCE>
          </dept>
          <employee>
       </employees>
</org>

now i want to replace the node REFERENCE with actual data at the XPath /org/depts/dept[2].
so the output xml should be like below.

    <org>
       <depts>
          <dept>
             <deptId>1009</deptId>
                <deptName>IT</deptName>
                <deptAccessCode>IT-1009</deptAccessCode>
          </dept>
          <dept>
             <deptId>2344</deptId>
             <deptName>BPO</deptName>
             <deptAccessCode>BP-2344</deptAccessCode>
          </dept>
          </depts>
          <employees>
             <employee>
             <name>abc</name>
             <dept>
             <deptId>2344</deptId>
             <deptName>BPO</deptName>
             <deptAccessCode>BP-2344</deptAccessCode>
          </dept>
          <employee>
       </employees>
    </org>

i have several REFERENCE nodes in different elements referencing to different xpaths across the xml tree, which i need to replace them with actual data.

<someWhereInTheXmlTree>
<sometag>
   <REFERENCE>
      <LocationXPath>some/reference[1]/to/a/node[3]/in/the[4]/same/xml</LocationXPath>
   </REFERENCE>
</sometag>
<someWhereInTheXmlTree>
...
<ffff>
<bbbb>
   <REFERENCE>
      <LocationXPath>abc/xyz[1]/node[4]/element</LocationXPath>
   </REFERENCE>
</bbbb>
<ffff>

please help me on this.
Thanks in advance for the help.

So far i have implemented one XSLT to replace the references but now i am facing unwanted empty name spaces.

Here is my XSLT

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:tib="http://www.tibco.com/bw/xslt/custom-functions"
xmlns="http://www.realestate.org/residential/2010/schemas" >

<xsl:output omit-xml-declaration="no" indent="yes" method = "xml" />
<xsl:strip-space elements="*"/>
<xsl:param name="myxml"  />
<xsl:template match="node()|@*">
    <xsl:param name="isNodeToReplace"><xsl:call-template name="ReferenceCheck" /></xsl:param>
    <xsl:choose>
        <xsl:when test="$isNodeToReplace='true'">
            <xsl:call-template name="replaceWithData">
                <xsl:with-param name="ref"><xsl:value-of select="." /></xsl:with-param>
            </xsl:call-template>
        </xsl:when>
        <xsl:otherwise>
            <xsl:copy>
                <xsl:apply-templates select="node()|@*"/>
            </xsl:copy>
        </xsl:otherwise>
    </xsl:choose>
</xsl:template>

<xsl:template name="ReferenceCheck">
    <xsl:choose>
        <xsl:when test="name(child::*[1])='REFERENCE' and name(child::*[1]//child::*[1])='LocationXPath'">true</xsl:when>
        <xsl:otherwise>false</xsl:otherwise>
    </xsl:choose>
</xsl:template>

<xsl:template name="replaceWithData">
    <xsl:param name="ref" />
    <xsl:copy-of select="tib:evaluate($myxml,$ref)" />
</xsl:template>

</xsl:stylesheet>

in the above XSLT i am passing the entire xml (same xml, which is being processed) as a parameter $myxml

Below is my sample input XML –this is just a snippet of xml,The actual xml file which i am dealing with is too large and contains so complex tree structure.How ever this sample xml is suffice enough to produce my problem.

Input file

<?xml version="1.0" encoding="UTF-8"?>
<org xmlns="http://www.realestate.org/residential/2010/schemas">
    <depts>
        <dept>
            <deptId>1</deptId>
            <deptName>health</deptName>
            <deptAccessCode>HL007845</deptAccessCode>
        </dept>
    </depts>
    <employees>
        <employee>
            <name>TOM</name>
            <dept>
                <REFERENCE>
                    <LocationXPath>/org/depts/dept[1]</LocationXPath>
                </REFERENCE>
            </dept>
        </employee>
    </employees>
</org>

my output file

<?xml version="1.0" encoding="UTF-8"?>
<org xmlns="http://www.realestate.org/residential/2010/schemas">
    <depts>
        <dept>
            <deptId>1</deptId>
            <deptName>health</deptName>
            <deptAccessCode>HL007845</deptAccessCode>
        </dept>
    </depts>
    <employees>
        <employee>
            <name>TOM</name>
            <dept xmlns="">
                <deptId>1</deptId>
                <deptName>health</deptName>
                <deptAccessCode>HL007845</deptAccessCode>
            </dept>
        </employee>
    </employees>
</org>  

Where as Expected output

<?xml version="1.0" encoding="UTF-8"?>
<org xmlns="http://www.realestate.org/residential/2010/schemas">
    <depts>
        <dept>
            <deptId>1</deptId>
            <deptName>health</deptName>
            <deptAccessCode>HL007845</deptAccessCode>
        </dept>
    </depts>
    <employees>
        <employee>
            <name>TOM</name>
            <dept>
                <deptId>1</deptId>
                <deptName>health</deptName>
                <deptAccessCode>HL007845</deptAccessCode>
            </dept>
        </employee>
    </employees>
</org>

so i am getting unwanted empty name space in << dept xmlns=””>> in the replaced root element.
Hope this could clearly explain my problem
Thanks in Advance

ultimately i have found the solution at the link below to remove the unwanted empty name spaces.
http://social.msdn.microsoft.com/forums/en-US/xmlandnetfx/thread/0de59291-ef3a-4a4c-9ca5-17923b16a504

Here is the new XSLT

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:tib="http://www.tibco.com/bw/xslt/custom-functions"
xmlns="http://www.realestate.org/residential/2010/schemas" >

<xsl:output omit-xml-declaration="no" indent="yes" method = "xml" />
<xsl:strip-space elements="*"/>
<xsl:param name="myxml"  />
<xsl:template match="node()|@*">
    <xsl:param name="isNodeToReplace"><xsl:call-template name="ReferenceCheck" /></xsl:param>
    <xsl:choose>
        <xsl:when test="$isNodeToReplace='true'">
            <xsl:call-template name="replaceWithData">
                <xsl:with-param name="ref"><xsl:value-of select="." /></xsl:with-param>
            </xsl:call-template>
        </xsl:when>
        <xsl:otherwise>
            <xsl:copy>
                <xsl:apply-templates select="node()|@*"/>
            </xsl:copy>
        </xsl:otherwise>
    </xsl:choose>
</xsl:template>

<xsl:template name="ReferenceCheck">
    <xsl:choose>
        <xsl:when test="name(child::*[1])='REFERENCE' and name(child::*[1]//child::*[1])='LocationXPath'">true</xsl:when>
        <xsl:otherwise>false</xsl:otherwise>
    </xsl:choose>
</xsl:template>

<xsl:template name="replaceWithData">
    <xsl:param name="ref" />
    <xsl:apply-templates select="tib:evaluate($myxml,$ref)" mode="move-to-namespace">
        <xsl:with-param name="namespace" select="'http://www.realestate.org/residential/2010/schemas'" />
    </xsl:apply-templates>
</xsl:template>
<xsl:template match="*" mode="move-to-namespace">
    <xsl:param name="namespace" />
    <xsl:element name="{local-name()}" namespace="{$namespace}">
        <xsl:copy-of select="@*" />
        <xsl:apply-templates select="node()" mode="move-to-namespace">
            <xsl:with-param name="namespace" select="$namespace"/>
        </xsl:apply-templates>
    </xsl:element>
</xsl:template>

<xsl:template match="text() | comment() | processing-instruction()" mode="move-to-namespace">
    <xsl:copy/>
</xsl:template>

</xsl:stylesheet>

if any xslt expert refines it further to avoid any unnecessary instruction with proper explanation its very glad.

Thanks in advance

Answer 1

In general:

• Not posiible in pure XSLT 1.0.

• Not possible in pure XSLT 2.0

• May be possible in pure XSLT 3.0 — read about the
  <xsl:evaluate> instruction.

• In XSLT 1.0 you may be lucky if your XSLT processor implements the EXSLT dyn:evaluate() extension function (a few do).
  Otherwize, you will have to write an extension function to select the
  nodes and return them back.

If there are restrictions on the syntax of the XPath expressions, then it may be possible to implement a pure XSLT 1.0 solution.