[SOLVED] XSLT to produce nested lists

Turbocapitalist · 01-29-2023, 06:27 AM

I have a question about how to use XSL (Extensible Stylesheet Language) to transform some basic XML into an unordered list while retaining the nesting. Here is a sample script containing the XML and the XSL which I have tried.

Code:

#!/usr/bin/python3

import lxml.etree as et

outline = '''<?xml version="1.0" encoding="UTF-8"?>
<xml>
 <o text="1">
  <o text="1.A" />
  <o text="1.B" />
  <o text="1.C" />
 </o>
 <o text="2">
  <o text="2.A">
   <o text="2.A.1" />
   <o text="2.A.2" />
  </o>
  <o text="2.B">
   <o text="2.B.1" />
   <o text="2.B.2" />
  </o>
  <o text="2.C" />
 </o>
</xml>
'''

xsl = '''
<xsl:stylesheet version="2.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
  <xsl:template match="/">
    <ul>
      <xsl:for-each select=".//o">
        <li>
          <xsl:value-of select="@text" />
        </li>
      </xsl:for-each>
    </ul>
  </xsl:template>
</xsl:stylesheet>
'''

dom = et.fromstring(outline.encode())
xslt = et.XML(xsl)
transform = et.XSLT(xslt)
dom2 = transform(dom)
html = et.tostring(dom2,pretty_print=True)
print(f"{html}")

That results in a flat list with a single level:

Code:

<ul>\n  <li>1</li>\n  <li>1.A</li>\n  <li>1.B</li>\n  <li>1.C</li>\n  <li>2</li>\n  <li>2.A</li>\n  <li>2.A.1</li>\n  <li>2.A.2</li>\n  <li>2.B</li>\n  <li>2.B.1</li>\n  <li>2.B.2</li>\n  <li>2.C</li>\n</ul>\n

I've tried a lot of variants but don't fully understand XSL and can't get past a flat list.

What I am hoping to figure out is how to preserve the nested aspects. The expected result would be this instead:

Code:

<ul>\n  <li>1\n    <ul>\n   <li>1.A</li>\n   <li>1.B</li>\n   <li>1.C</li>\n   </ul>\n </li>\n   <li>2\n    <ul>\n   <li>2.A\n    <ul>\n   <li>2.A.1</li>\n   <li>2.A.2</li>\n   </ul>\n </li>\n   <li>2.B\n    <ul>\n   <li>2.B.1</li>\n   <li>2.B.2</li>\n   </ul>\n </li>\n   <li>2.C</li>\n   </ul>\n</li>\n   </ul>\n

How should the XSL look to achieve that?

ntubski · 01-29-2023, 09:25 AM

The ".//o" expression produces a flat list of all o elements; rather than iterating over the list, you want to apply <xsl:template match> recursively:

Code:

<xsl:stylesheet version="2.0" 
                xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>

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

  <xsl:template match="o">
    <li>
      <xsl:value-of select="@text" />
      <xsl:apply-templates/>
    </li>
  </xsl:template>
</xsl:stylesheet>

Turbocapitalist · 01-29-2023, 09:32 AM

Thanks, I'll now try experimenting with that. Would there be a way for to include the UL elements too each time a new level is entered? The goal is that the output is valid HTML. I suppose some kind of XPath with first-child or last-child would be too much of a kludge.

ntubski · 01-29-2023, 10:09 AM

Quote:

Originally Posted by Turbocapitalist

Would there be a way for to include the UL elements too each time a new level is entered? The goal is that the output is valid HTML.

Oh, missed that part. I guess every level but the last?

Code:

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

  <xsl:template match="o[./o]"> <!-- o elements containing o elements -->
      <ul>
        <li>
          <xsl:value-of select="@text" />
          <xsl:apply-templates/>
        </li>
      </ul>
  </xsl:template>

  <xsl:template match="o">      <!-- other o elements (i.e., lowest level) -->
    <li>
      <xsl:value-of select="@text" />
    </li>
  </xsl:template>
</xsl:stylesheet>

Turbocapitalist · 01-29-2023, 01:57 PM

Thanks! Having two XPath patterns was the right approach. The "o[./o]" was needed to wrap the descendants inside unordered lists:

Code:

xsl = '''                                                                       
<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:template match="/">                                                      
    <ul>                                                                        
      <xsl:apply-templates/>                                                    
    </ul>                                                                       
  </xsl:template>                                                               
                                                                                
  <xsl:template match="o[./o]"> <!-- o elements containing o elements -->       
      <li>                                                                      
          <xsl:value-of select="@text" />                                       
        <ul>                                                                    
          <xsl:apply-templates/>                                                
        </ul>                                                                   
      </li>                                                                     
  </xsl:template>                                                               
                                                                                
  <xsl:template match="o">      <!-- other o elements (i.e., lowest level) -->  
    <li>                                                                        
      <xsl:value-of select="@text" />                                           
    </li>                                                                       
  </xsl:template>                                                               
</xsl:stylesheet>                                                               
'''

I'm familiar with XPaths but the use of XSLT is still (hours) new to me. So this helped a lot.