Question

鉴于此输入XML：

<?xml version="1.0" encoding="ISO-8859-1" ?>
<agrisResources xmlns:ags="http://purl.org/agmes/1.1/" xmlns:dc="http://purl.org/dc/elements/1.1/">
    <agrisResource bibliographicLevel="AM" ags:ARN="^aSF17^b00003">
        <dc:subject xml:lang="en">Penaeidae</dc:subject>
        <dc:subject xml:lang="en">Vibrio harveyi</dc:subject>
        <dc:subject xml:lang="en">Vibrio parahaemolyticus</dc:subject>
        <dc:subject>
            <ags:subjectClassification scheme="ags:ASC">ASFA-1</ags:subjectClassification>
            <ags:subjectThesaurus xml:lang="en" scheme="ags:ASFAT">Bacterial diseases</ags:subjectThesaurus>
            <ags:subjectThesaurus xml:lang="en" scheme="ags:ASFAT">Fish diseases</ags:subjectThesaurus>
            <ags:subjectThesaurus xml:lang="en" scheme="ags:ASFAT">Genes</ags:subjectThesaurus>
        </dc:subject>
    </agrisResource>
</agrisResources>

我想将具有相同属性的项目分组，因此输出将如下所示：

<dc:subject xml:lang="en">Penaeidae||Vibrio harveyi||Vibrio parahaemolyticus</dc:subject>
<dc:subject>
    <ags:subjectClassification scheme="ags:ASC">ASFA-1</ags:subjectClassification>
    <ags:subjectThesaurus xml:lang="en" scheme="ags:ASFAT">Bacterial diseases||Fish diseases||Genes</ags:subjectThesaurus>
</dc:subject>

基本上，我的分组规则是在节点有多个值的情况下组合节点的值，例如dc:subject和ags:subjectThesaurus。我在标题中指定具有相同属性的值，因为我不确定是否可以按标记对它们进行分组而不指定其属性来区分它们。

换句话说，区分

<dc:subject>Penaeidae</dc:subject>

这

<dc:subject>
    <ags:subjectThesaurus>Bacterial diseases</ags:subjectThesaurus>
</dc:subject>

更新

INPUT XML

<?xml version="1.0" encoding="ISO-8859-1" ?>
<agrisResources xmlns:ags="http://purl.org/agmes/1.1/" xmlns:dc="http://purl.org/dc/elements/1.1/">
    <agrisResource bibliographicLevel="AM" ags:ARN="^aSF17^b00003">
        <dc:creator>
            <ags:creatorPersonal>Doe, John</ags:creatorPersonal>
            <ags:creatorPersonal>Smith, Jason T.</ags:creatorPersonal>
            <ags:creatorPersonal>Doe, Jane E.</ags:creatorPersonal>
        </dc:creator>
        <dc:subject xml:lang="en">Penaeidae</dc:subject>
        <dc:subject xml:lang="en">Vibrio harveyi</dc:subject>
        <dc:subject xml:lang="en">Vibrio parahaemolyticus</dc:subject>
        <dc:subject>
            <ags:subjectClassification scheme="ags:ASC">ASFA-1</ags:subjectClassification>
            <ags:subjectThesaurus xml:lang="en" scheme="ags:ASFAT">Bacterial diseases</ags:subjectThesaurus>
            <ags:subjectThesaurus xml:lang="en" scheme="ags:ASFAT">Fish diseases</ags:subjectThesaurus>
            <ags:subjectThesaurus xml:lang="en" scheme="ags:ASFAT">Genes</ags:subjectThesaurus>
        </dc:subject>
    </agrisResource>
</agrisResources>

所需输出

分组规则：使用双管道||组合值作为重复元素的分隔符，例如<ags:creatorPersonal>，<dc:subject xml:lang="en">和<ags:subjectThesaurus xml:lang="en" scheme="ags:ASFAT">。将其他元素保留为不符合该规则的元素。

<?xml version="1.0" encoding="ISO-8859-1" ?>
<agrisResources xmlns:ags="http://purl.org/agmes/1.1/" xmlns:dc="http://purl.org/dc/elements/1.1/">
    <agrisResource bibliographicLevel="AM" ags:ARN="^aSF17^b00003">
        <dc:creator>
            <ags:creatorPersonal>Doe, John||Smith, Jason T.||Doe, Jane E.</ags:creatorPersonal>
        </dc:creator>
        <dc:subject xml:lang="en">Penaeidae||Vibrio harveyi||Vibrio parahaemolyticus</dc:subject>
        <dc:subject>
            <ags:subjectClassification scheme="ags:ASC">ASFA-1</ags:subjectClassification>
            <ags:subjectThesaurus xml:lang="en" scheme="ags:ASFAT">Bacterial diseases||Fish diseases||Genes</ags:subjectThesaurus>
        </dc:subject>
    </agrisResource>
</agrisResources>

以下是基于此answer的代码：

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:dc="http://purl.org/dc/terms/"
            xmlns:ags="http://purl.org/agmes/1.1/"
            xmlns:agls="http://www.naa.gov.au/recordkeeping/gov_online/agls/1.2"
            xmlns:dcterms="http://purl.org/dc/terms/">
    <xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>
    <xsl:strip-space elements="*"/>

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

    <xsl:template match="ags:subjectThesaurus|dc:subject">
        <xsl:copy>
            <xsl:apply-templates select="@* | text()"/>
                <xsl:call-template name="NextSibling"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="ags:subjectThesaurus[@scheme = preceding-sibling::*[1][self::ags:subjectThesaurus]/@scheme]|dc:subject[@xml:lang = preceding-sibling::*[1][self::dc:subject]/@xml:lang]"/>

    <xsl:template match="ags:subjectThesaurus|dc:subject" mode="includeSib">
        <xsl:value-of select="concat('||', .)"/>
            <xsl:call-template name="NextSibling"/>
        </xsl:template>

    <xsl:template name="NextSibling">
        <xsl:apply-templates select="following-sibling::*[1][self::ags:subjectThesaurus and @scheme = current()/@scheme]|following-sibling::*[1][self::dc:subject and @xml:lang = current()/@xml:lang]" mode="includeSib"/>
    </xsl:template>
</xsl:stylesheet>

我唯一的问题是它只是转换ags:subjectThesaurus而不是dc:subject节点。我的输出如下：

<dc:subject xml:lang="en">Penaeidae</dc:subject>
<dc:subject xml:lang="en">Vibrio harveyi</dc:subject>
<dc:subject xml:lang="en">Vibrio parahaemolyticus</dc:subject>
<dc:subject>
    <ags:subjectThesaurus xml:lang="en" scheme="ags:ASFAT">Bacterial diseases||Fish diseases||Genes</ags:subjectThesaurus>
</dc:subject>

如何修改我的代码，以便它还将dc:subject节点与xml:lang属性相同？

修改

根据michael.hor257k的建议以及answer使用Muenchian方法的建议，以下是我尝试的内容：

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:dc="http://purl.org/dc/terms/"
            xmlns:ags="http://purl.org/agmes/1.1/"
            xmlns:agls="http://www.naa.gov.au/recordkeeping/gov_online/agls/1.2"
            xmlns:dcterms="http://purl.org/dc/terms/">
    <xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>
    <xsl:strip-space elements="*"/>
    <xsl:key name="kNodeSubject" match="dc:subject[@xml:lang]" use="@xml:lang"/>
    <xsl:key name="subjectThesaurus" match="dc:subject/ags:subjectThesaurus" use="@scheme"/>
    <xsl:template match="node() | @*">
        <xsl:copy>
            <xsl:apply-templates select="node() | @*"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="dc:subject[generate-id() = generate-id(key('kNodeSubject', @xml:lang)[1])]">
        <xsl:copy>
            <xsl:apply-templates select="@*"/>
            <xsl:apply-templates select="key('kNodeSubject', @xml:lang)" mode="concat"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="dc:subject/ags:subjectThesaurus[generate-id() = generate-id(key('subjectThesaurus', @scheme)[1])]">
        <xsl:copy>
            <xsl:apply-templates select="@*"/>
            <xsl:apply-templates select="key('subjectThesaurus', @scheme)" mode="concat"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="dc:subject|subjectThesaurus" mode="concat">
        <xsl:value-of select="."/>
            <xsl:if test="position() != last()">
                <xsl:text>||</xsl:text>
            </xsl:if>
    </xsl:template>

    <xsl:template match="dc:subject"/>
    <xsl:template match="ags:subjectThesaurus"/>
</xsl:stylesheet>

当我应用上面的代码时，节点ags:subjectThesaurus消失了，<dc:subject xml:lang="en">的值也没有分组。我不知道我是否匹配正确，我使用match="dc:subject[@xml:lang]"作为<xsl:key name="kNodeSubject"，因为节点ags:subjectThesaurus是<dc:subject>的孩子。

提前致谢。

Answer 1

考虑以下示例：

<强> XML

<root xmlns:dc="http://purl.org/dc/terms/" xmlns:ags="http://purl.org/agmes/1.1/">
  <dc:subject xml:lang="en">Penaeidae</dc:subject>
  <dc:subject xml:lang="en">Vibrio harveyi</dc:subject>
  <dc:subject xml:lang="fr">Franca premier</dc:subject>
  <dc:subject xml:lang="fr">Franca deux</dc:subject>
  <dc:subject xml:lang="en">Vibrio parahaemolyticus</dc:subject>
  <dc:subject>
    <ags:subjectThesaurus xml:lang="en" scheme="ags:ASFAT">Bacterial diseases</ags:subjectThesaurus>
    <ags:subjectThesaurus xml:lang="en" scheme="ags:ASFAT">Fish diseases</ags:subjectThesaurus>
    <ags:subjectThesaurus xml:lang="en" scheme="ags:ASFAT">Genes</ags:subjectThesaurus>
    <ags:subjectThesaurus xml:lang="en" scheme="ags:B">Bees</ags:subjectThesaurus>
    <ags:subjectThesaurus xml:lang="en" scheme="ags:B">Birds</ags:subjectThesaurus>
  </dc:subject>
</root>

XSLT 1.0

<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:dc="http://purl.org/dc/terms/"
xmlns:ags="http://purl.org/agmes/1.1/">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>

<xsl:key name="subj-by-lang" match="dc:subject[@xml:lang]" use="@xml:lang"/>
<xsl:key name="thes-by-scheme" match="ags:subjectThesaurus" use="@scheme"/>

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

<xsl:template match="root">
    <xsl:copy>
        <!-- group subjects by lang -->
        <xsl:for-each select="dc:subject[@xml:lang][count(. | key('subj-by-lang', @xml:lang)[1]) = 1]">
             <dc:subject xml:lang="{@xml:lang}">
                <xsl:for-each select="key('subj-by-lang', @xml:lang)">
                    <xsl:value-of select="."/>
                    <xsl:if test="position() != last()">
                        <xsl:text>||</xsl:text>
                    </xsl:if>
                </xsl:for-each>
             </dc:subject>  
        </xsl:for-each>
        <!-- process other nodes -->
        <xsl:apply-templates select="node()[not(self::dc:subject[@xml:lang])]"/>
    </xsl:copy>
</xsl:template>

<xsl:template match="dc:subject">
    <xsl:copy>
        <!-- group thesauri by scheme    -->
        <xsl:for-each select="ags:subjectThesaurus[count(. | key('thes-by-scheme', @scheme)[1]) = 1]">
             <dc:subjectThesaurus xml:lang="{@xml:lang}" scheme="{@scheme}">
                <xsl:for-each select="key('thes-by-scheme', @scheme)">
                    <xsl:value-of select="."/>
                    <xsl:if test="position() != last()">
                        <xsl:text>||</xsl:text>
                    </xsl:if>
                </xsl:for-each>
             </dc:subjectThesaurus> 
        </xsl:for-each>
    </xsl:copy>
</xsl:template>

</xsl:stylesheet>

<强>结果

<?xml version="1.0" encoding="UTF-8"?>
<root xmlns:dc="http://purl.org/dc/terms/" xmlns:ags="http://purl.org/agmes/1.1/">
  <dc:subject xml:lang="en">Penaeidae||Vibrio harveyi||Vibrio parahaemolyticus</dc:subject>
  <dc:subject xml:lang="fr">Franca premier||Franca deux</dc:subject>
  <dc:subject>
    <dc:subjectThesaurus xml:lang="en" scheme="ags:ASFAT">Bacterial diseases||Fish diseases||Genes</dc:subjectThesaurus>
    <dc:subjectThesaurus xml:lang="en" scheme="ags:B">Bees||Birds</dc:subjectThesaurus>
  </dc:subject>
</root>

加了：

根据您的澄清，我怀疑您想要做一些更简单的事情：只需将一些叶子节点（即没有子元素的节点）连接在一起，并将其他节点保留原样。

这是加入dc:subject中agrisResource个叶节点的示例：

<xsl:template match="agrisResource">
    <xsl:copy>
        <!-- join subjects with no children -->
        <dc:subject>
            <!-- copy the attributes of the first subject with no children -->
            <xsl:copy-of select="dc:subject[not(*)][1]/@*"/>
            <!-- concat the values of all subjects with any attributes -->
            <xsl:for-each select="dc:subject[not(*)]">
                <xsl:value-of select="."/>
                <xsl:if test="position() != last()">
                    <xsl:text>||</xsl:text>
                </xsl:if>
            </xsl:for-each>
         </dc:subject>  
        <!-- process other nodes -->
        <xsl:apply-templates select="node()[not(self::dc:subject[not(*)])]"/>
    </xsl:copy>
</xsl:template>

这可以通过使用基于元素名称的密钥来推广。

使用xslt-1.0对具有相同属性的值进行分组

1 个答案:

加了：