如何将平面xml数据转换为分层数据xml [英] How to convert flat xml data to hierarchical data xml

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问题描述

我必须将平面 xml 转换为分层 xml.我不知道这个任务.以下是转换输入.

I have to convert a flat xml to hierarchical xml. I have no idea for this task. Below is input for conversion.

输入:-

<body>
    <p class="title">Article Title</p>
    <p class="Authors">abc, pqr and xyz</p>
    <p class="intro">here is introdution text......</p>        
    <p class="head1">1: Heading level 1</p>
    <p>some text here</p>
    <p>some text here</p>
    <p class="head2">1.1: Heading  level 2</p>
    <p>some text here</p>
    <p>some text here</p>        
    <p class="head3">1.1.1: Heading  level 3</p>
    <p>some text here</p>
    <p>some text here</p>        
    <p class="head1">2: Heading level 1</p>
    <p class="head2">2.1: Heading  level 2</p>
    <p>some text here</p>
    <p>some text here</p>        
    <p class="head3">2.1.1: Heading  level 3</p>
    <p>some text here</p>
    <p>some text here</p>        
    <p class="head3">2.1.2: Heading  level 3</p>
    <p>some text here</p>
    <p>some text here</p>        
</body>

我想将其转换为下面给定的 xml.

I want to convert it to below given xml.

输出:-

<article>
    <title>Article Title</title>
    <Authors>abc, pqr and xyz</Authors>
    <introduction>here is introdution text......</introduction>
    <body>
        <sec level="1">
            <title>1: Heading level 1</title>
            <p>some text here</p>
            <p>some text here</p>
            <sec level="2">
                <title>1.1: Heading  level 2</title>
                <p>some text here</p>
                <p>some text here</p>
                <sec level="3">
                    <title>1.1.1: Heading  level 3</title>
                    <p>some text here</p>
                    <p>some text here</p>                    
                </sec>
            </sec>
        </sec>
        <sec level="1">
            <title>2: Heading level 1</title>
            <sec level="2">                
                <title>2.1: Heading  level 2</title>
                <p>some text here</p>
                <p>some text here</p>
                <sec level="3">
                    <title>2.1.1: Heading  level 3</title>
                    <p>some text here</p>
                    <p>some text here</p>                        
                </sec>
                <sec level="3">
                    <title>2.1.2: Heading  level 3</title>
                    <p>some text here</p>
                    <p>some text here</p>                        
                </sec>
            </sec>                
        </sec>
    </body>
</article>

我不知道使用 xslt 转换它.

I havn't any idea to convert it using xslt.

我低于输出:-

   <article>
  <p class="title">Article Title</p>
  <p class="Authors">abc, pqr and xyz</p>
  <p class="intro">here is introdution text......</p>
  <body>
     <p class="head1">1: Heading level 1</p>
     <p>some text here</p>
     <p>some text here</p>
     <p class="head2">1.1: Heading level 2</p>
     <p>some text here</p>
     <p>some text here</p>
     <p class="head3">1.1.1: Heading level 3</p>
     <p>some text here</p>
     <p>some text here</p>
     <p class="head1">2: Heading level 1</p>
     <p class="head2">2.1: Heading level 2</p>
     <p>some text here</p>
     <p>some text here</p>
     <p class="head3">2.1.1: Heading level 3</p>
     <p>some text here</p>
     <p>some text here</p>
     <p class="head3">2.1.2: Heading level 3</p>
     <p>some text here</p>
     <p>some text here</p>
  </body>
  </article>

我的代码是:-

      <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
     xmlns:xs="http://www.w3.org/2001/XMLSchema" exclude-result-prefixes="xs" version="2.0">
     <xsl:output method="xml" indent="yes"/>
     <xsl:strip-space elements="*"/>

     <xsl:template match="/">
        <xsl:apply-templates/>
     </xsl:template>

     <xsl:template match="@* | node()">
        <xsl:copy copy-namespaces="no">
           <xsl:apply-templates select="@* | node()"/>
        </xsl:copy>
     </xsl:template>
     <xsl:template match="html">
        <xsl:apply-templates/>
     </xsl:template>

     <xsl:template match="head"/>
     <xsl:template match="body">
        <xsl:element name="article">
           <xsl:apply-templates select="p[@class='title']|p[@class='Authors']|p[@class='intro']"/>
           <xsl:element name="body">
              <xsl:apply-templates select="p[preceding-sibling::p[@class='intro']]"/>
           </xsl:element>
        </xsl:element>
     </xsl:template>
  </xsl:stylesheet>

推荐答案

您可以使用以下 xslt 代码获得预期输出:

You can use the below xslt code for expected output:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:mf="http://example.com/mf"
exclude-result-prefixes="xs mf" version="2.0">

<xsl:output indent="yes"/>

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

<xsl:function name="mf:group" as="element(section)*">
    <xsl:param name="entries" as="element(p)*"/>
    <xsl:param name="level" as="xs:integer"/>
    <xsl:for-each-group select="$entries"
        group-starting-with="p[@class = concat('head',$level)]">
        <xsl:variable name="P_ID" select="generate-id(.)"/>
        <section name="{@class}">
            <title>
                <xsl:value-of select="."/>
            </title>
            <xsl:if test="following-sibling::p[1][not(@class)]">
                <ps>
                    <xsl:apply-templates
                        select="following-sibling::p[not(@class)][generate-id(preceding-sibling::p[@class][1]) = $P_ID]"
                    />
                </ps>
            </xsl:if>

            <xsl:sequence select="mf:group(current-group() except ., ($level + 1))"/>
        </section>
    </xsl:for-each-group>
</xsl:function>

<xsl:template match="body">
    <xsl:copy>
        <xsl:sequence select="mf:group(p[contains(@class,'head')], 1)"/>
    </xsl:copy>
</xsl:template>

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