堆栈内存溢出
  简体   繁体   English

使用xslt-1.0将具有相同属性的值分组

[英]Grouping values with same attributes using xslt-1.0

 原文  2017-08-14 03:32:31  7   1    xml/ xslt/ xpath/ xslt-1.0

问题描述

Given this input XML: 给定此输入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>

I would like to group items with the same attributes, so the output would be like this: 我想对具有相同属性的项目进行分组，因此输出如下所示： 

<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>

Basically, my rule for the grouping is to combine the values of the nodes if that node have multiple values, eg dc:subject , and ags:subjectThesaurus . 基本上，如果该节点具有多个值（例如dc:subject和ags:subjectThesaurus ，则分组的规则是合并节点的值。 I specify in my title to group values with same attributes because I'm not really sure if it is possible to just group them by their tags without specifying their attributes to differentiate them. 我在标题中指定了对具有相同属性的值进行分组的方法，因为我不确定是否可以仅按其标签对它们进行分组而不指定其属性来区分它们。

In other words, differentiate 换句话说，区分 

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

from 从 

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

UPDATE UPDATE

INPUT XML 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: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>

Desired Output 期望的输出

Rules on grouping: Combine the values using double pipe || 分组规则：使用双管道||组合值 as separator for repeating elements, eg <ags:creatorPersonal> , <dc:subject xml:lang="en"> and <ags:subjectThesaurus xml:lang="en" scheme="ags:ASFAT"> . 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"> 。 Leave other elements as is that does not meet that rule. 保留不符合该规则的其他元素。 

<?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>

Below is my code based from this answer : 下面是我基于此答案的代码： 

<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>

My only problem is that it is only transforming the ags:subjectThesaurus but not the dc:subject node. 我唯一的问题是，它仅转换ags:subjectThesaurus而不转换dc:subject节点。 My output looks like this: 我的输出如下所示： 

<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>

How can I modify my code such that it will also group the dc:subject node with the same xml:lang attribute? 如何修改我的代码，使其也将具有相同xml:lang属性的dc:subject节点分组？

EDIT 编辑

Based on the suggestion of michael.hor257k and from this answer to use the Muenchian method, below is what I tried: 基于michael.hor257k的建议，并根据使用Muenchian方法的答案 ，以下是我尝试的方法： 

<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>

When I applied the code above, the nodes ags:subjectThesaurus are gone and the values of <dc:subject xml:lang="en"> are not grouped either. 当我应用上面的代码时，节点ags:subjectThesaurus消失了，并且<dc:subject xml:lang="en">也未分组。 I don't know if I have the match right, I used the match="dc:subject[@xml:lang]" for the <xsl:key name="kNodeSubject" because the node ags:subjectThesaurus is the child of <dc:subject> . 我不知道我是否具有匹配权，我对<xsl:key name="kNodeSubject"使用match="dc:subject[@xml:lang]" ，因为节点ags:subjectThesaurus是<dc:subject> 。

Thanks in advance. 提前致谢。

1 个解决方案

解决方案1
 1 已采纳  2017-08-17 04:55:36

Consider the following example: 考虑以下示例：

XML 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 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>

Result 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>

Added: 添加：

Based on your clarifications, I suspect you want to do something much simpler: just join together some leaf nodes (ie nodes with no child elements) and leave the others as is. 根据您的澄清，我怀疑您想做一些更简单的事情：仅将一些叶子节点（即没有子元素的节点）连接在一起，而将其他叶子节点保持原样。

Here's an example joining the dc:subject leaf nodes within agrisResource : 这是在agrisResource加入dc:subject叶子节点的agrisResource ： 

<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>

This could be generalized by using a key based on an element's name. 这可以通过使用基于元素名称的键来概括。

暂无
暂无

声明:本站的技术帖子网页，遵循CC BY-SA 4.0协议，如果您需要转载，请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

相关问题 有人可以帮助我将其从 XSLT-2.0 转换为 XSLT-1.0 吗？ *更新*，试图创建 Muenchian Grouping XSLT-1.0 - Can someone assist me in converting this from XSLT-2.0 to XSLT-1.0? *Update*, attempted to create Muenchian Grouping XSLT-1.0 XSLT-1.0：将逗号分隔的值转换为元素值 - XSLT-1.0: Convert comma-separated values to element values 使用 XSLT-1.0 按第一个 0 值分组的 CSV 到 XML - CSV to XML using XSLT-1.0 group by first 0 value 如何拆分字符串<br>在 xml 中使用 xslt-1.0 - How to split a string with <br> in xml using xslt-1.0 如何在XSLT-1.0中连接XML标签？ - How to concat XML tags in XSLT-1.0? 参考 xslt-1.0 中的当前 XML 文档 - Refer to current XML document in xslt-1.0 使用Muenchian分组xslt 1.0 - Using Muenchian grouping xslt 1.0 使用XSLT 1.0对输出进行分组 - Grouping output using XSLT 1.0 使用XSLT 1.0进行复杂分组 - Complex Grouping using XSLT 1.0 如何在xslt-1.0中使用SOH和ETX - how to use SOH and ETX in xslt-1.0
相关标签
 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM