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我正在尝试将 XML 文档转换为一个列表，其中的值基于前面兄弟姐妹的属性。

示例 XML：

<myRoot>
    <Person>Craig</Person>
    <Person rank="10">Woody</Person>
    <Person>Brian</Person>
    <Person>Michael</Person>    
    <Person rank="20">Emily</Person>
    <Person>Chris</Person>
</myRoot>

我想要的是：

<myNewRoot>
    <Index>1: Craig</Index>
    <Index>10: Woody</Index>
    <Index>11: Brian</Index>
    <Index>12: Michael</Index>
    <Index>20: Emily</Index>
    <Index>21: Chris</Index>
</myNewRoot>

我被卡住了，无法确定使用 @rank 属性的最后一个先前兄弟与当前节点之间的距离。

这是我当前的样式表

<xsl:template match="Person">
    <xsl:element name="Index">
        <xsl:choose>

            <xsl:when test="./@rank">
                <xsl:value-of select="./@rank"/>
            </xsl:when>


            <xsl:when test="preceding-sibling::Person[@rank]">
                <xsl:value-of select="count(.|preceding-sibling::*[. &gt; current()/preceding-sibling::Person[@rank][1]])   +       preceding-sibling::Person[@rank][1]/@rank"/>

            </xsl:when>


            <xsl:otherwise>
                <xsl:value-of select="position()"/>
            </xsl:otherwise>

        </xsl:choose>

        <xsl:text>: </xsl:text>
        <xsl:value-of select="."/>
    </xsl:element>
</xsl:template>

我只是无法让 count() 功能正常工作并继续以

<myNewRoot>
    <Index>1: Craig</Index>
    <Index>10: Woody</Index>
    <Index>11: Brian</Index>
    <Index>11: Michael</Index>
    <Index>20: Emily</Index>
    <Index>21: Chris</Index>
</myNewRoot>
4

1 回答 1

2

一种解决方案是使用递归模板根据打印的先前值获取当前值，该值作为参数传递给模板。

<xsl:output method="xml" indent="yes" />

<xsl:template match="myRoot">
    <myNewRoot>
        <xsl:call-template name="make-index" />
    </myNewRoot>
</xsl:template>

<xsl:template name="make-index">
    <xsl:param name="element" select="Person" />
    <xsl:param name="count" select="'0'" />

    <!-- Continue if there is some element left -->
    <xsl:if test="$element">
        <!-- Obtain number to be printed next -->
        <xsl:variable name="next-rank">
            <xsl:choose>
                <!-- If the rank attribute is present, output its value -->
                <xsl:when test="$element[1]/@rank">
                    <xsl:value-of select="$element[1]/@rank" />
                </xsl:when>
                <!-- If the rank attribute is not present, increase the previous
                     printed value by one -->
                <xsl:otherwise>
                    <xsl:value-of select="$count + 1" />
                </xsl:otherwise>                    
            </xsl:choose>
        </xsl:variable>
        <!-- Output the index along with the value of the current index -->
        <Index>
            <xsl:value-of select="concat($next-rank, ': ',  $element[1])" />
        </Index>
        <!-- Recurse until we do not have any element left -->
        <xsl:call-template name="make-index">
            <xsl:with-param name="element" select="$element[position() > 1]" />
            <xsl:with-param name="count" select="$next-rank" />
        </xsl:call-template>
    </xsl:if>
</xsl:template>

更新。下面的解决方案不依赖递归，可能不如前一个有效（在这个有更复杂的 XPath 操作），但更短并且依赖于对兄弟进行分组，这是与前一个不同的方法。

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

    <xsl:output method="xml" indent="yes" />

    <xsl:template match="myRoot">
        <myNewRoot>
            <!-- Print all the elements before the first element with a rank
                 attribute defined -->
            <xsl:apply-templates select="Person[(preceding-sibling::Person[@rank])][not(@rank)]" mode="print" />
            <!-- Match all the elements with a rank attribute defined -->
            <xsl:apply-templates select="Person[@rank]" />
        </myNewRoot>
    </xsl:template>

    <!-- Match the set of Person elements with @rank defined -->
    <xsl:template match="Person[@rank]">
        <!-- Obtain id of current node before losing the context -->
        <xsl:variable name="id" select="generate-id()" />
        <!-- Match the current node along all the following siblings without @rank such
             as their nearest preceding-sibling with @rank defined is the current
             element, i.e all the elements between the current element and the next
             element with @rank defined -->
        <xsl:apply-templates select=".|following-sibling::Person[not(@rank)][generate-id(preceding-sibling::Person[@rank][1]) = $id]" mode="print">
            <xsl:with-param name="rank" select="@rank" />
        </xsl:apply-templates>
    </xsl:template>

    <!-- Print the information from a Person node, using rank to determine
         the position -->
    <xsl:template match="Person" mode="print">
        <xsl:param name="rank" select="'1'" />
        <Index>
            <xsl:value-of select="concat($rank + position() - 1, ': ', .)" />
        </Index>
    </xsl:template>

注意：我假设您使用 XSLT 1.0 的两种解决方案。如果您使用的是 XSLT 2.0，该解决方案将比以前的解决方案更容易。

于 2013-02-21T19:36:57.107 回答