[xsl] Seek an XPath expression that removes trailing spaces

["Roger L Costello costello@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

Hi Folks,

I want to convert this:

<Airport_Name>LOWELL FLD                    </Airport_Name>

to this:

<name>LOWELL FLD</name>

The value of the <name> element must be identical to the value of the
<Airport_Name> element, except without the trailing spaces.

Multiple internal consecutive spaces, if present, must be preserved. Hence,
the normalize-space() function cannot be used.

I have a solution, but it requires a mix of XPath and a user-defined XSLT
function. I seek a pure XPath solution.

Here is the approach I took:

Convert the value of Airport_Name to a sequence of codepoints:

let $cp := string-to-codepoints(string(Airport_Name)) return

Reverse the sequence of codepoints:

let $rev := reverse($cp) return

Get the index of the first non-blank character:

let $idx := f:index-of-first-non-blank-char($rev,1) return

Extract the subsequence starting at the index:

let $subseq := $rev[position() ge $idx] return

Reverse the subsequence:

let $rev2 := reverse($subseq) return

Convert the codepoints to a string and return the string:

codepoints-to-string($rev2)

Obviously, f:index-of-first-non-blank-char() is not XPath; it's a user-defined
XSLT function that I created:

<xsl:function name="f:index-of-first-non-blank-char" as="xs:integer">
    <xsl:param name="str" as="xs:integer+"/>
    <xsl:param name="idx" as="xs:integer"/>
    <xsl:choose>
        <xsl:when test="$str[1] ne 32">  <!-- 32 = decimal value of space char
-->
            <xsl:sequence select="$idx"/>
        </xsl:when>
        <xsl:otherwise>
            <xsl:sequence
select="f:index-of-first-non-blank-char($str[position() gt 1], $idx + 1)"/>
        </xsl:otherwise>
    </xsl:choose>
</xsl:function>

There must be a better (simpler, shorter, more straightforward) way to solve
this problem.

Help, please!

/Roger