XSLT - 获取最接近当前时间戳的节点

Question

我有一组带有时间戳格式日期的节点，我正在使用<xsl:for-each>进行迭代。如何获得最接近保存在$currentDate中的当前日期的节点？

 <xsl:variable name="currentDate" select="1396332574" />
 <items>
   <item>
      <tmstmp>1374249600</tmstmp>
   </item>
   <item>
      <tmstmp>1374249600</tmstmp>
   </item>
   <item>
      <tmstmp>1374257700</tmstmp>
   </item>
   <item>
      <tmstmp>1374418800</tmstmp>
   </item>
   <item>
      <tmstmp>2777068800</tmstmp>
   </item>
 </items>

Answer 1

试试

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    version="1.0">
    <xsl:strip-space elements="*"/>
    <xsl:output omit-xml-declaration="yes" indent="yes"/>
    <xsl:variable name="currentDate" select="1396332574" />
    <xsl:template match="/">
        <xsl:for-each select="items/item">
            <xsl:sort select="$currentDate - ." data-type="number" order="ascending"/>
            <xsl:if test="position() = 1">
                <xsl:copy-of select="."/>
            </xsl:if>
        </xsl:for-each>
    </xsl:template>
</xsl:stylesheet>

它输出：

<item>
   <tmstmp>1374418800</tmstmp>
</item>