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copied!<p><strong>Edited</strong>: Per clarification from the original poster, the first string delimited by the underscores should not have its starting letter capitalized.</p> <p><strong>This solution does not use recursion and should be quite more efficient</strong>.</p> <p><strong>Here is the new solution</strong>:</p> <pre><code>string-join( (for $i in 1 to count(tokenize(.,'_')), $s in tokenize(.,'_')[$i], $fl in substring($s,1,1), $tail in substring($s,2) return if($i eq 1) then $s else concat(upper-case($fl), $tail) ), '' ) </code></pre> <p>The result is now exactly as required:</p> <pre><code>underLinedString </code></pre> <p><strong>Below is the old solution</strong>.</p> <p><strong>The replace() function supposes that you have a fixed replacement</strong> -- therefore it is not the best tool for solving this problem.</p> <p><strong>Here is a one-liner XPath 2.0 solution</strong> (and it certainly can be used as part of an XSLT 2.0 transformation :)     ):</p> <pre><code>string-join( (for $s in tokenize(.,'_'), $fl in substring($s,1,1), $tail in substring($s,2) return concat(upper-case($fl), $tail) ), '' ) </code></pre> <p><strong>When we use the above expression in an XSLT 2.0 transformation like this</strong>:</p> <pre><code><xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xs="http://www.w3.org/2001/XMLSchema"> <xsl:template match="/"> <xsl:sequence select= "string-join( (for $s in tokenize(.,'_'), $fl in substring($s,1,1), $tail in substring($s,2) return concat(upper-case($fl), $tail) ), '' ) "/> </xsl:template> </xsl:stylesheet> </code></pre> <p>and apply it on this XML document:</p> <pre><code><t>under_lined_String</t> </code></pre> <p><strong>the wanted result is produced</strong>:</p> <pre><code>UnderLinedString </code></pre>

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