input xml to output xml transformation using xslt

Question

I have a input and desired output xml file , but not able to write generic xsl transformer for that. Can anybody here help me out ? address/billing may have more elements at runtime, which should be directly copied to main person block.

<searchPersonResponse>
 <persons>
  <person>
   <name>name2</name>
   <address>
   <billing>
     <city>xx</city>
     <state>yyyy</state>
     <zip>zzzzz</zip>
   </billing>
   </address>
  </person>
  <person>
    <name>name1</name>
    <address>
      <billing>
       <city>xx</city>
       <state>yyyy</state>
       <zip>zzzzz</zip>
      </billing>
    </address>
   </person>
  </persons>
 </searchPersonResponse>

desired output xml

<searchPersonResponse>
 <persons>
  <person>
     <name>name2</name>
     <city>xx</city>
     <state>yyyy</state>
     <zip>zzzzz</zip>
  </person>
  <person>
       <name>name1</name>
       <city>xx</city>
       <state>yyyy</state>
       <zip>zzzzz</zip>
   </person>
  </persons>
 </searchPersonResponse>

TddOrBust · Accepted Answer · 2013-02-15 13:12:05Z

2

EDITED: OP is only looking to "unwrap" address and billing elements when address is present with a billing child. XSLT now does that. Also, OP mentions needs a XSLT 1.0 solution; no XSLT 2.0 features were being used, so I simply changed the version to "1.0".

This XSLT:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    version="1.0">
    <xsl:template match="@*|node()">
        <xsl:copy>
                <xsl:apply-templates/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="address[child::billing]|billing[parent::address]">
        <xsl:apply-templates/>
    </xsl:template>
</xsl:stylesheet>

When applied to this XML:

<searchPersonResponse>
    <persons>
        <person>
            <name>name2</name>
            <address>
                <billing>
                <city>xx</city>
                <state>yyyy</state>
                <zip>zzzzz</zip>
                </billing>
            </address>
        </person>
        <person>
            <name>name1</name>
            <address>
                <billing>
                <city>xx</city>
                <state>yyyy</state>
                <zip>zzzzz</zip>
                </billing>
            </address>
        </person>
    </persons>
</searchPersonResponse>

Produces the desired result:

<searchPersonResponse>
    <persons>
        <person>
            <name>name2</name>
            <city>xx</city>
            <state>yyyy</state>
            <zip>zzzzz</zip>
        </person>
        <person>
            <name>name1</name>
            <city>xx</city>
            <state>yyyy</state>
            <zip>zzzzz</zip>
        </person>
    </persons>
</searchPersonResponse>

The identity template will copy all nodes and attributes. The address and billing matching template (which matches either element) will copy their children but not themselves.

edited Feb 15, 2013 at 13:12

answered Feb 15, 2013 at 3:51

TddOrBust

4143 silver badges12 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

Daniel Haley Over a year ago

You should consolidate your two templates by changing the match to match="address|billing" (+1)

TddOrBust Over a year ago

Thank you! That is a good call. The perils of rushing to be the first to answer. :) Thanks!

amarzeet Over a year ago

Steveo, I want a exact match of address/billing and when this is found inside elements should be copied directly. Solution, you gave is looking for 'or' condition. To make it 'and' , I changed match to match='address/billing', but it is not working

amarzeet Over a year ago

other thing, I would like to add is , I am using version="1.0" .

TddOrBust Over a year ago

@amarzeet, see my updated answer. The "address/billing" doesn't work because you are only selecting billing elements that are children of address elements, so the address element gets copied by the identity template. My updated answer matches on address elements with billing children and billing elements with address parents. Also, I wasn't using anything specific to XSLT 2.0, so changing the version number in the xsl:stylesheet element was all that was needed to make it usable by 1.0 processors.

Collectives™ on Stack Overflow

input xml to output xml transformation using xslt

1 Answer 1

5 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

5 Comments

Your Answer

Sign up or log in

Post as a guest

Related