How to merge xml nodes with common attribute value

Question

I would like to take some XML document and merge all nodes on some level who have the same attribute value. For instance, if I have a document of this form,

<xml>
 <books id="X">
  <book name="Story"/>
  <book name="Tale"/>
 </books>
 <books id="X">
  <book name="Folklore"/>
  <book name="Magic"/>
 </books>
 <books id="Y">
  <book name="Harry Potter"/>
  <book name="LotR"/>
 </books>
</xml>

then I would like to be able to convert it to the following document:

<xml>
 <books id="X">
  <book name="Story"/>
  <book name="Tale"/>
  <book name="Folklore"/>
  <book name="Magic"/>
 </books>
 <books id="Y">
  <book name="Harry Potter"/>
  <book name="LotR"/>
 </books>
</xml>

If at all possible, I would very much like to multiple solutions. For instance both one using XSLT and one using some other language.

You cannot modify XML source with XPath. You should check XSLT instead. Also note that question that contain no attempts to the solve issue should be closed as *too broad* or *off-topic* — Andersson, Sep 18 '18 at 07:32
This is a grouping issue. If you are using XSLT 1.0, then read up on [Muenchian Grouping](http://www.jenitennison.com/xslt/grouping/muenchian.html). If you can use XSLT 2.0 or above, then [xsl:for-each-group](https://www.xml.com/pub/a/2003/11/05/tr.html) will be your friend. If you can't get it to work, post what you have tried, and I am sure you will get an answer. Thanks! — Tim C, Sep 18 '18 at 07:53

score 0 · Answer 1 · answered Sep 18 '18 at 08:26

<xsl:template match="@*|node()">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="books[1][@id='X']">
        <xsl:copy>
            <xsl:apply-templates select="@*"/>
        <xsl:apply-templates/>
            <xsl:apply-templates select="following-sibling::books[@id='X']/book"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="books[@id='X' and preceding-sibling::books[@id='X']]"/>
    <xsl:template match="book">
        <xsl:copy>
        <xsl:apply-templates select="@*"/>
            <xsl:apply-templates/>
        </xsl:copy>
    </xsl:template>
check it.

score 0 · Accepted Answer · answered Sep 18 '18 at 08:34

you can try this:

<xsl:template match="xml">
        <xsl:for-each-group select="books" group-by="@id">
            <xsl:copy>
                <xsl:attribute name="id" select="current-grouping-key()"/>
                <xsl:apply-templates select="current-group()/*"/>
            </xsl:copy>
        </xsl:for-each-group>
    </xsl:template>

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

See conversion at https://xsltfiddle.liberty-development.net/gWmuiJR

How to merge xml nodes with common attribute value

2 Answers2