Removing a parent element in xml while keeping it's children using xslt

Question

I want to transform the following xml,

<pets>
    <Pet>
        <category>
            <id>4</id>
            <name>Lions</name>
        </category>
        <id>9</id>
        <name>Lion 3</name>
        <photoUrls>
            <photoUrl>url1</photoUrl>
            <photoUrl>url2</photoUrl>
        </photoUrls>
        <status>available</status>
        <tags>
            <tag>
                <id>1</id>
                <name>tag3</name>
            </tag>
            <tag>
                <id>2</id>
                <name>tag4</name>
            </tag>
        </tags>
    </Pet>
</pets>

in to this xml format.

<pets>
    <Pet>
        <category>
            <id>4</id>
            <name>Lions</name>
        </category>
        <id>9</id>
        <name>Lion 3</name>
        <photoUrl>url1</photoUrl>
        <photoUrl>url2</photoUrl>
        <status>available</status>
        <tag>
            <id>1</id>
            <name>tag3</name>
        </tag>
        <tag>
            <id>2</id>
            <name>tag4</name>
        </tag>
    </Pet>
</pets>

I tried to write a template as follows, but it removes the parent element with it's children.

<xsl:template match="photoUrls"/>

How can this be done in xslt. Any help is appreciated.

This seems to work. , Still any feedback is appreciated. Is this the best way or are there other better ways of doing this? — Ravindra Ranwala, Mar 23 '16 at 23:57

score 5 · Answer 1 · answered Mar 24 '16 at 05:14

5

I would do it this way :

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output indent="yes" method="xml" />

    <xsl:template match="photoUrls|tags">
       <!-- Apply identity transform on child elements of photoUrls/tags-->
        <xsl:apply-templates select="*"/>
    </xsl:template>

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

answered Mar 24 '16 at 05:14

har07

88,338
12
84
137

This works when I don't have any namespaces in my xml but fails to remove the tags when I have namespaces (soapenv) in my xml. How can I modify it to work? – Prabakaran Thodithot Sembiyan Nov 21 '17 at 11:18
Got it, I added [local-name()] before the tag and it works. – Prabakaran Thodithot Sembiyan Nov 21 '17 at 11:19
@PrabakaranThodithotSembiyan Would you mind sharing the updated code? – DbxD Jun 04 '19 at 11:27

score 3 · Answer 2 · answered Mar 24 '16 at 00:04

The following xslt does the job,

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output indent="yes" method="xml" />

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

    <xsl:template match="photoUrls">
        <xsl:copy-of select="photoUrl" />
    </xsl:template>

    <xsl:template match="tags">
        <xsl:copy-of select="tag" />
    </xsl:template>
</xsl:stylesheet>

But if you have any other way of doing this please don't hesitate to post your answer here.

Hikmat · Answer 3 · 2016-03-24T18:11:21.080

0

You can use this code also.

<?xml version="1.0" encoding="UTF-8" ?><xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:output method="xml"/>

<xsl:template match="photoUrls|tags">
    <xsl:copy-of select="node()"/>
</xsl:template>
<xsl:template match="@*|node()">
    <xsl:copy>
        <xsl:apply-templates select="@*|node()"/>
    </xsl:copy>
</xsl:template>
</xsl:transform>

edited Mar 24 '16 at 18:11

answered Mar 24 '16 at 06:30

Hikmat

450
4
19

Don't copy attributes if you're not copying the parent element. – michael.hor257k Mar 24 '16 at 11:08

Removing a parent element in xml while keeping it's children using xslt

3 Answers3