XSLT获得最后一个元素 [英] XSLT getting last element

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本文介绍了XSLT获得最后一个元素的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我试图在xml中找到最后一个元素,

I am trying to find the last element in my xml, which looks like

    <list>
        <element name="A" />
        <element name="B" >
            <element name="C" />
            <element name="D" >
                <element name="D" />
                <element name="E" />
                <element name="F" />
                <element name="G" />
            </element>
        <element name="H" />
        <element name="I" />
    </list>

我需要获取某种反向菜单,其中当前元素和父级突出显示为活动",而同级元素突出为不活动".取而代之的是,只有当我单击"D"元素时,我才会有一棵凌乱的树.

I need to get some kind of reverse menu, where current element and parents are highlighted as "active" and sibling as "inactive". Instead in result I have a messy tree only when I suppose "D" element clicked.

双重D元素是我的问题.当我使用 第一次选择首次出现 D元素时(调试器显示),select="//element[@name='D'][last()]"select="//element[@name='D' and last()]"(正确的是吗?).这是xsl

Double D elements are my problem. When I use select="//element[@name='D'][last()]" or select="//element[@name='D' and last()]" (btw which one is correct?) first time first occurrence of D element is selected (debugger shows that). Here is xsl

<xsl:template match="list">
    <xsl:apply-templates select="//navelement[@name = 'D'][last()]" mode="active"/>
</xsl:template>

<xsl:template match="element">
    <ul class="menu">
    <xsl:apply-templates select="preceding-sibling::node()" mode="inactive"/>
        <li><a>....</a></li>
    <xsl:apply-templates select="following-sibling::node()" mode="inactive"/>
    </ul>   
    <xsl:apply-templates select="parent::element" mode="active"/>
</xsl:template>

<xsl:template match="element" mode="inactive">
        <li><a>....</a></li>
</xsl:template>

推荐答案

您需要将last()索引放在节点列表结果上,而不是作为选择标准的一部分.试试

You need to put the last() indexing on the nodelist result, rather than as part of the selection criteria. Try

(//element[@name='D'])[last()]

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