Get all attribute values from XML

I have the following XML sample file.

<root>
<ul id="nav">
    <li id="id1">some</li>
    <li id="id2"> unimportant
        </li>
    <li id="id3"> text.
        </li>
</ul>

<ul id="tag">
    <li id="id1">SOME</li>
    <li id="id2"> MORE
        </li>
    <li id="id3"> TEXT.
        </li>
</ul>
</root>

I would like to use xmllint to extract all attribute values of all <li> elements which are listed under <ul id="nav">.

I can use

xmllint --xpath '//ul[@id="nav"]/li/@id' sample.xml

which will return

id="id1" id="id2" id="id3"

However, I would like to get

id1 id2 id3

I tried

xmllint --xpath 'string(//ul[@id="nav"]/li/@id)' sample.xml

but this only return the value of the first attribute

id1

Can I extract all values without resorting to other tools like awk, sed etc.? They may be delimited by either space or newline, it does not really matter in my case.

Solution

If you can use xmlstarlet, you can use the -v (--value-of) command line option:

xmlstarlet sel -t -v "//ul[@id='nav']/li/@id" sample.xml

I don't think you can use HTML as an input for the sel command, but I could be wrong.

I know you can with the tr (translate) command though (with --html). Then you could use an XSLT...

xmlstarlet tr --html sample.xsl sample.xml

XSLT (sample.xsl)

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="text"/>
  <xsl:strip-space elements="*"/>

  <xsl:template match="/">
    <xsl:for-each select="//ul[@id='nav']/li/@id">
      <xsl:value-of select="concat(.,'&#xA;')"/>
    </xsl:for-each>
  </xsl:template>

</xsl:stylesheet>