How to get the following sibling in XSLT

xml, xpath, xslt

Solution

This is actually a completely XPath question.

Use:

/*/project[title = 'Project X']/following-sibling::project[1]

This selects any first following sibling `Project` of any `Project` element that is a child of the top element in the XML document and the string value of at least of one of its `title` children is the string `"Project X"`.

XSLT - based verification:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

 <xsl:template match="/">
     <xsl:copy-of select=
      "/*/project[title = 'Project X']/following-sibling::project[1]"/>
 </xsl:template>
</xsl:stylesheet>

When this transformation is applied on the provided XML document:

<projects>
    <project>
        <number>1</number>
        <title>Project X</title>
    </project>
    <project>
        <number>2</number>
        <title>Project Y</title>
    </project>
    <project>
        <number>3</number>
        <title>Project Z</title>
    </project>
</projects>

the XPath expression is evaluated and the correctly-selected element is copied to the output:

<project>
   <number>2</number>
   <title>Project Y</title>
</project>

Problem

I am fairly new to XSLT and this is my XML: ``` <projects> <project> <number>1</number> <title>Project X</title> </project> <project> <number>2</number> <title>Project Y</title> </project> <project> <number>3</number> <title>Project Z</title> </project> </projects> ``` If I have one project and want to get the sibling that follows it, how can I do that? This code doesn't seem to work for me: ``` /projects[title="Project X"]/following-sibling ```

Original source