Question

I have an XML of this kind:

<xml>
    <node1>1.1</node1>
    <node1>1.2</node1>
    <node1>1.3</node1>

    <node2>2.1</node2>
    <node2>2.2</node2>
    <node2>2.3</node2>

    <node3>3.1</node3>
    <node3>3.2</node3>
    <node3>3.3</node3>
</xml>

and I want to get the following output:
line: 1.1 + 2.1 + 3.1
line: 1.2 + 2.2 + 3.2
line: 1.3 + 2.3 + 3.3

Is there a way I can iterate over these nodes simultaneously and keep track of my current position in each of three lists or do I have to wrap these items into a bigger block and iterate over blocks?

I’m using XSL 1.0.

Answer 1

As easy as this:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output method="text"/>
 <xsl:strip-space elements="*"/>

 <xsl:template match="node1">
   <xsl:variable name="vPos" select="position()"/>
     <xsl:value-of select=". + ../node2[$vPos] + ../node3[$vPos]"/>
     <xsl:text>&#xA;</xsl:text>
 </xsl:template>
 <xsl:template match="text()"/>
</xsl:stylesheet>

When this transformation is applied on the following XML document (based on the provided one):

<xml>
    <node1>1</node1>
    <node1>2</node1>
    <node1>3</node1>

    <node2>4</node2>
    <node2>5</node2>
    <node2>6</node2>

    <node3>7</node3>
    <node3>8</node3>
    <node3>9</node3>
</xml>

the wanted, correct result is produced:

12
15
18

Do note:

1. In XSLT 1.0 and XSLT 2.0 one can use the FXSL template/function zip-with3().

2. In XPath 3.0 (XSLT 3.0) there will be a standard function map-pairs(), but there is no map-tripples() standard function. One can use this function to produce an intermediate result and then use it again to produce the final result.

3. As noted by Ian Roberts, the presence of xsl:strip-space in this solution is important — without it the position() function produces different results and the transformation doesn’t perform as required.