使用XSLT排序的XML到XML

Question

我有幾個巨大的XML，我只需要對其中的一小部分進行排序。 作為輸出，我應該具有相同的XML，但具有排序的子結構。 這是一個例子：

<testStructure>
<parentStruct>
    <firstpPreChild>some value here</firstpPreChild>
    <secondPreChild>some other value</secondPreChild>
    <thirdPreChild>third value here</thirdPreChild>
    <fourtPreChild>fourth value here</fourtPreChild>
    <struct id="5">
        <num>5</num>
    </struct>
    <struct id="4">
        <num>4</num>
    </struct>
    <struct id="1">
        <num>1</num>
    </struct>
    <struct id="2">
        <num>2</num>
    </struct>
    <struct id="3">
        <num>3</num>
    </struct>
     <firstAdditionalChild>some value here</firstAdditionalChild>
    <secondAdditionalChild>some other value</secondAdditionalChild>
    <thirdAdditionalChild>third value here</thirdAdditionalChild>
    <fourtAdditionalChild>fourth value here</fourtAdditionalChild>-->
</parentStruct>
<otherStruct>
    <firstChild>some value here</firstChild>
    <secondChild>some other value</secondChild>
    <thirdChild>third value here</thirdChild>
    <fourtChild>fourth value here</fourtChild>
</otherStruct>

應該轉化為

<testStructure>
<parentStruct>
    <firstpPreChild>some value here</firstpPreChild>
    <secondPreChild>some other value</secondPreChild>
    <thirdPreChild>third value here</thirdPreChild>
    <fourtPreChild>fourth value here</fourtPreChild>
    <struct id="1">
        <num>1</num>
    </struct>
    <struct id="2">
        <num>2</num>
    </struct>
    <struct id="3">
        <num>3</num>
    </struct>
    <struct id="4">
        <num>4</num>
    </struct>
    <struct id="5">
        <num>5</num>
    </struct>
     <firstAdditionalChild>some value here</firstAdditionalChild>
    <secondAdditionalChild>some other value</secondAdditionalChild>
    <thirdAdditionalChild>third value here</thirdAdditionalChild>
    <fourtAdditionalChild>fourth value here</fourtAdditionalChild>-->
</parentStruct>
<otherStruct>
    <firstChild>some value here</firstChild>
    <secondChild>some other value</secondChild>
    <thirdChild>third value here</thirdChild>
    <fourtChild>fourth value here</fourtChild>
</otherStruct>

作為排序標准，可以使用num或@id 。 我試過這樣的變體：

<xsl:template match="node()|@*">
<xsl:copy>
   <xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>

可以，但是將排序后的結構從其原始位置移開。 不幸的是，我需要與輸出相同的結構。

先謝謝您的幫助！

Answer 1

XSLT 1.0樣式表按相鄰分組然后排序：

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:key name="kStructByFirstPreceding"
             match="struct"
             use="generate-id(
                     preceding-sibling::struct[
                        not(preceding-sibling::*[1]/self::struct)
                     ][1]
                  )"/>
    <xsl:template match="node()|@*" name="identity">
        <xsl:copy>
            <xsl:apply-templates select="node()|@*"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="struct[not(preceding-sibling::*[1]/self::struct)]">
        <xsl:apply-templates select=".|key('kStructByFirstPreceding',
                                           generate-id())"
                             mode="copy">
            <xsl:sort select="@id"/>
        </xsl:apply-templates>
    </xsl:template>
    <xsl:template match="struct"/>
    <xsl:template match="node()" mode="copy">
        <xsl:call-template name="identity"/>
    </xsl:template>
</xsl:stylesheet>

輸出：

<testStructure>
    <parentStruct>
        <firstpPreChild>some value here</firstpPreChild>
        <secondPreChild>some other value</secondPreChild>
        <thirdPreChild>third value here</thirdPreChild>
        <fourtPreChild>fourth value here</fourtPreChild>
        <struct id="1">
            <num>1</num>
        </struct>
        <struct id="2">
            <num>2</num>
        </struct>
        <struct id="3">
            <num>3</num>
        </struct>
        <struct id="4">
            <num>4</num>
        </struct>
        <struct id="5">
            <num>5</num>
        </struct>
        <firstAdditionalChild>some value here</firstAdditionalChild>
        <secondAdditionalChild>some other value</secondAdditionalChild>
        <thirdAdditionalChild>third value here</thirdAdditionalChild>
        <fourtAdditionalChild>fourth value here</fourtAdditionalChild>--&gt; 
    </parentStruct>
    <otherStruct>
        <firstChild>some value here</firstChild>
        <secondChild>some other value</secondChild>
        <thirdChild>third value here</thirdChild>
        <fourtChild>fourth value here</fourtChild>
    </otherStruct>
</testStructure>

更簡單的XSLT 2.0解決方案：

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:template match="node()|@*">
        <xsl:copy>
            <xsl:apply-templates select="node()|@*"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="*[struct]">
        <xsl:copy>
            <xsl:for-each-group select="*"
                                group-adjacent="boolean(self::struct)">
                <xsl:apply-templates select="current-group()">
                    <xsl:sort select="@id[current-grouping-key()]"/>
                </xsl:apply-templates>
            </xsl:for-each-group>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>