How to flatten a very complex XML into a new XML containing all nodes at root level

自古美人都是妖i 提交于 2019-12-12 23:57:14

问题


I am currently trying to flatten a large recurisve XML document, so that all the nested elements stay on root level but get an additional new attribute ("parent_id=...") to still keep the relations between the nodes.

Each node has a lot of sub-nodes which i also need to grab, so the content has to stay the same.

The file is very large (500k Rows - 33 MB in Size)

Example XML:

<product-catalog ...>
  <category id="1">
    <content>
      ...
     </content>
     <category id="2">
        <content>
        ...
        </content>
     </category>
     <category id="3">
        <content>
        ...
        </content>
        <category id="4">
           ...
        </category>
        <category id="5">
           ...
        </category>
     </category>
   </category>
</product-catalog>

Required flattened output:

<product-catalog>
  <category id="1" parent_id="0">
     <content>...</content>
  </category>
  <category id="2" parent_id="1">
     <content>...</content>
  </category>
  <category id="3" parent_id="1">
     <content>...</content>
  </category>
  <category id="4" parent_id="3">
     <content>...</content>
  </category>
  <category id="5" parent_id="3">
     <content>...</content>
  </category>
</product-catalog>

tried this so far, but it only delivers root category (not really an xslt-expert... ;))

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

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

    <xsl:template match="category">  
        <xsl:element name="category">
            <xsl:apply-templates select="@* | node() [not(child::category)]"/>      
        </xsl:element>
    </xsl:template>


    <!-- remove -->
    <xsl:template match="translations" />


</xsl:stylesheet>

回答1:


Consider the following example:

XML

<product-catalog>
    <category id="1">
        <content>A1</content>
        <category id="2">
            <content>B</content>
        </category>
        <category id="3">
            <content>C1</content>
            <content>C2</content>
            <category id="4">
                <content>D</content>
            </category>
            <category id="5">
                <content>E</content>
            </category>
        </category>
        <content>A2</content>
    </category>
</product-catalog>

XSLT 1.0

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

<xsl:template match="/product-catalog">
    <xsl:copy>
        <xsl:apply-templates select="category"/>
    </xsl:copy>
</xsl:template>

<xsl:template match="category">
    <category id="{@id}" parent_id="{parent::category/@id}">
        <xsl:copy-of select="content"/>
    </category>
    <xsl:apply-templates select="category"/>
</xsl:template>

</xsl:stylesheet>

Result

<?xml version="1.0" encoding="UTF-8"?>
<product-catalog>
  <category id="1" parent_id="">
    <content>A1</content>
    <content>A2</content>
  </category>
  <category id="2" parent_id="1">
    <content>B</content>
  </category>
  <category id="3" parent_id="1">
    <content>C1</content>
    <content>C2</content>
  </category>
  <category id="4" parent_id="3">
    <content>D</content>
  </category>
  <category id="5" parent_id="3">
    <content>E</content>
  </category>
</product-catalog>

how could i copy all existing attributes of <category...> and add only parent_id

Try:

<xsl:template match="category">
    <category parent_id="{parent::category/@id}">
        <xsl:copy-of select="@* | content"/>
    </category>
    <xsl:apply-templates select="category"/>
</xsl:template>


来源:https://stackoverflow.com/questions/57445013/how-to-flatten-a-very-complex-xml-into-a-new-xml-containing-all-nodes-at-root-le

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