XSLT for Variable Element Name for multiple same elements

Question

I have a piece of XML data with information for multiple persons represented as -

<phoneContact>
    <firstName>XXXXX</firstName>
    <middleName>Y</middleName>
    <lastName>ZZZZZ</lastName>
    <phone>1234567890</phone>
</phoneContact>
<phoneContact>
    <firstName>AAAA</firstName>
    <middleName>B</middleName>
    <lastName>CCCCC</lastName>
    <phone>9876543210</phone>
</phoneContact>

There may be any number of persons available. I want to transform this into -

<phoneContact1>
    <firstName>XXXXX</firstName>
    <middleName>Y</middleName>
    <lastName>ZZZZZ</lastName>
    <phone>1234567890</phone>
</phoneContact1>
<phoneContact2>
    <firstName>AAAA</firstName>
    <middleName>B</middleName>
    <lastName>CCCCC</lastName>
    <phone>9876543210</phone>
</phoneContact2>

.. and so forth. How do I construct an XSL for-each code that creates multiple such different element names ?

Thank you for any help with this.

Are you sure you want to do this? It's not a good structure to have, as it makes it very difficult to select all phoneContacts for processing. Normally, you would use an attribute to distinguish an element from its siblings of same name. — michael.hor257k
– michael.hor257k, Commented Mar 1, 2016 at 18:10
Agreed. Your requested output XML form is a much poorer design than your input XML. What's your bigger picture goal? — kjhughes
– kjhughes, Commented Mar 1, 2016 at 18:17
Thank you for your response. Agree, that is a better structure. But I want to do this transform for another implementation downstream, where a DB table will have one record per person with the possibility of having multiple phone numbers. — mrigs13
– mrigs13, Commented Mar 1, 2016 at 18:44

wero · Accepted Answer · 2016-03-01 18:09:42Z

1

Inside the loop over the phoneContact elements you can use <xsl:element> and the the position function to create your numbered phoneContacts:

<xsl:for-each select="phoneContact"> 
    <xsl:element name="phoneContact{position()}">
     ...
   </xsl:element>
</xsl:for-each>

answered Mar 1, 2016 at 18:09

wero

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1 Comment

mrigs13 Over a year ago

And that worked perfectly, kind Sir. THANK YOU. Here is what I implemented finally -

<xsl:for-each select="phoneContact">     <xsl:element name="phoneContact{position()}">         <firstName><xsl:value-of select="firstName"/></firstName>         <middleName><xsl:value-of select="middleName"/></middleName>         <lastName><xsl:value-of select="lastName"/></lastName>         <phone><xsl:value-of select="phone"/></phone>     </xsl:element> </xsl:for-each>

Rama Krishnam · Accepted Answer · 2016-03-02 09:44:21Z

0

we can implement using identity function as below

<xsl:template match="node()|@*">
    <xsl:copy>
        <xsl:apply-templates select="@*|node()"/>
    </xsl:copy>
</xsl:template>
  <xsl:template match="/a/phoneContact">
<xsl:element name="phoneContact{position()}">

  <xsl:apply-templates select="@*|node()"/>

</xsl:element>
  </xsl:template>

answered Mar 2, 2016 at 9:44

Rama Krishnam

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