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copied!<p><em>The previous two answers seem to be not exact</em>.</p> <p><strong>Here is one possible solution</strong>:</p> <p>You want to find the first node with the maximum value for the following function:</p> <pre><code>100*number(@text='Login') +10*number(@href='Login.php') + 1*number(@index=0) </code></pre> <p><strong>In XPath 2.0</strong> this can be expressed as a single XPath expression in the following way:</p> <pre><code> /*/link[ 100*number(@text='Login') +10*number(@href='Login.php') + 1*number(@index=0) eq max(/*/link /(100*number(@text='Login') +10*number(@href='Login.php') + 1*number(@index='0') ) ) ] </code></pre> <p><strong>In XPath 1.0 constructing such a one-lener expression would be extremely difficult</strong>, if possible at all, and even if possible, such an XPath expression will be impossible to understand, prove correct and/or maintain.</p> <p>However, <em>selecting the best-matching <code>link</code> element is possible within any language that is a host of XPath 1.0</em>. </p> <p><strong>One example below is with XSLT 1.0</strong> as the hosting language:</p> <pre><code><xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output omit-xml-declaration="yes" indent="yes"/> <xsl:template match="/"> <xsl:for-each select="*/link"> <xsl:sort data-type="number" order="descending" select= "100*(@text='Login') +10*(@href='Login.php') + 1*(@index=0) "/> <xsl:if test="position() = 1"> <xsl:copy-of select="."/> </xsl:if> </xsl:for-each> </xsl:template> </xsl:stylesheet> </code></pre> <p><strong>when the above transformation is applied on this XML document</strong>:</p> <pre><code><links> <link name="A" text="Sign in" href="Login.php" index="0"/> <link name="B" text="Login" href="SignIn.php" index="15"/> <link name="C" text="Login" href="Login.php" index="22"/> </links> </code></pre> <p><strong>the correct result is produced</strong>:</p> <pre><code><link name="C" text="Login" href="Login.php" index="22" /> </code></pre> <p><strong>This reminds me of <a href="http://www.biglist.com/lists/xsl-list/archives/200207/msg00804.html" rel="nofollow noreferrer">another "Single XPath expression finding the best matches" problem</a></strong> <strong>I solved some seven year ago :)</strong></p>

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