XSL / XPath expression to check if a node contains at least one non-empty child

Question

I need to check if an XML node has at least one non-empty child. Applied to this XML the expression should return true

Answer 1

Here's one XPath that should accomplish the job:

count(/*/node/*[text()]) > 0

When used in a sample XSLT:

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

  <xsl:template match="/">
     <xsl:value-of select="count(/*/node/*[text()]) &gt; 0" />
  </xsl:template>

</xsl:stylesheet>

...which is, in turn, applied to the provided example XML:

<xml>
  <node>
    <node1/>
    <node2/>
    <node3>value</node3>
  </node>
</xml>

...the expected result is produced:

true

If we apply the same XSLT against a simply modified XML:

<xml>
  <node>
    <node1/>
    <node2/>
    <node3/>
  </node>
</xml>

...again, the expected result is produced:

false

Explanation:

The XPath used searches for all children of a <node> element (which are, in turn, children of the root element) that have a non-empty text value (as specified by text()); should the count of such <node> children be greater than 0, then the XPath resolves to true.

Answer 2

You just need <xsl:if test="/xml/node/* != ''" />.

In XPath an = or != comparison where one side is a node set and the other side is a string succeeds if any of the nodes in the set passes the comparison. Thus

not(x = '')

means "it is not the case that any x child element of the current node has an empty string value", which is fundamentally different from

x != ''

which means "at least one x child element of the current node has a string value that is not empty". In particular, if you want to check that all x children are empty, you need to use a "double-negative" test

not(x != '')

Answer 3

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
    <xsl:output omit-xml-declaration="yes" indent="yes"/>
    <xsl:strip-space elements="*"/>
    <xsl:template match="/">
        <xsl:value-of select="/*/node/*[string-length(text()) &gt;0]!=''"/>
    </xsl:template>
</xsl:stylesheet>

Explanation This will find the first node with a string length greater than zero and then compares such node contents with empty string (the comparison will return the existence of a non-empty string node); this code can also be used to look for a specific criteria in any node, for example identify the existence of a node which contains specific string or starts with some character or any other condition; please use this as the inner condition of the node reference for the code to work its magic.

Answer 4

More accurate, simpler and more efficient (no need to use the count() function):

  /*/node/*[text()]

If you want to exclude any element that has only whitespace-only text children, use:

  /*/node/*[normalize-space()]