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Home/ Questions/Q 6932099
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T11:39:47+00:00 2026-05-27T11:39:47+00:00

Given this XML/HTML: <dl> <dt>Label1</dt><dd>Value1</dd> <dt>Label2</dt><dd>Value2</dd> <dt>Label3</dt><dd>Value3a</dd><dd>Value3b</dd> <dt>Label4</dt><dd>Value4</dd> </dl> I want to find all

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Given this XML/HTML:

<dl>
  <dt>Label1</dt><dd>Value1</dd>
  <dt>Label2</dt><dd>Value2</dd>
  <dt>Label3</dt><dd>Value3a</dd><dd>Value3b</dd>
  <dt>Label4</dt><dd>Value4</dd>
</dl>

I want to find all <dt> and then, for each, find the following <dd> up until the next <dt>.

Using Ruby’s Nokogiri I am able to accomplish this like so:

dl.xpath('dt').each do |dt|
  ct  = dt.xpath('count(following-sibling::dt)')
  dds = dt.xpath("following-sibling::dd[count(following-sibling::dt)=#{ct}]")
  puts "#{dt.text}: #{dds.map(&:text).join(', ')}"
end
#=> Label1: Value1
#=> Label2: Value2
#=> Label3: Value3a, Value3b
#=> Label4: Value4

However, as you can see I’m creating a variable in Ruby and then composing an XPath using it. How can I write a single XPath expression that does the equivalent?

I guessed at:

following-sibling::dd[count(following-sibling::dt)=count(self/following-sibling::dt)]

but apparently I don’t understand what self means there.

This question is similar to XPath : select all following siblings until another sibling except there is no unique identifier for the ‘stop’ node.

This question is almost the same as xpath to find all following sibling adjacent nodes up til another type except that I’m asking for an XPath-only solution.

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  1. Editorial Team
    Editorial Team
    2026-05-27T11:39:47+00:00Added an answer on May 27, 2026 at 11:39 am

    One possible solution:

    dl.xpath('dt').each_with_index do |dt, i|
      dds = dt.xpath("following-sibling::dd[not(../dt[#{i + 2}]) or " +
                         "following-sibling::dt[1]=../dt[#{i + 2}]]")
      puts "#{dt.text}: #{dds.map(&:text).join(', ')}"
    end
    

    This relies on a value comparison of dt elements and will fail when there are duplicates. The following (much more complicated) expression does not depend on unique dt values:

    following-sibling::dd[not(../dt[$n]) or 
        (following-sibling::dt[1] and count(following-sibling::dt[1]|../dt[$n])=1)]
    

    Note: Your use of self fails because you’re not properly using it as an axis (self::). Also, self always contains just the context node, so it would refer to each dd inspected by the expression, not back to the original dt

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