XSLT Concatenate element values - xslt-2.0

Question

My requirement is to concatenate values coming at run time in the ref element. I could get separate values for each element and the concatenate it but still i will miss out text in between.
I have edited my questions as per suggestion made by Martin.
Here is my input XML:
<?xml version="1.0" encoding="UTF-8"?><rootelement>
<nref ref="f1"/> <nref ref="f2"/> <nref ref="f3"/>
<nref ref="f4"/> <nref ref="f5"/>
<pnotes>
<pnote id="f1">
<p>Some Text.</p>
</pnote>
<pnote id="f2">
<p><ref><year>1812</year>, <vol>3</vol><series>F. &
F.</series><pages>22</pages></ref>, at p. 27.</p>
</pnote>
<pnote id="f3">
<p><ref>[<year>1914</year>],
<vol>2</vol><series>A.B.C.</series><pages>94</pages></ref>.</p>
</pnote>
<pnote id="f4">
<p><ref><year>1955</year>,
<vol>2</vol><series>A.B.C</series><pages>509</pages></ref>.</p>
</pnote>
<pnote id="f5">
<p><ref>[<year>1805</year>], <series>A.C.</series><pages>21</pages> </ref>.</p>
</pnote>
</pnotes></rootelement>
I have tried following XSLT
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="xs"
version="2.0">
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="nref">
<xsl:variable name="ref" select="#ref"/>
<noteref>
<xsl:for-each select="//pnotes/pnote">
<xsl:if test="#id = $ref">
<xsl:value-of select="."/>
</xsl:if>
</xsl:for-each>
</noteref>
</xsl:template>
<xsl:template match="//pnotes[not(pnotes)]"/>
</xsl:stylesheet>
The result i get is not correct.
I need spaces in between the text.
Result :
<?xml version="1.0" encoding="UTF-8"?>
<rootelement>
<noteref> Some Text. </noteref>
<noteref> 1812, 3F. & F.22, at p. 27. </noteref>
<noteref> [1914], 2A.B.C.94. </noteref>
<noteref> 1955, 2A.B.C509. </noteref>
<noteref> [1805], A.C.21. </noteref>
</rootelement>
Required Result is:
<?xml version="1.0" encoding="UTF-8"?>
<rootelement>
<noteref> Some Text. </noteref>
<noteref> 1812, 3 F. & F. 22, at p. 27. </noteref>
<noteref> [1914], 2 A.B.C. 94. </noteref>
<noteref> 1955, 2 A.B.C 509. </noteref>
<noteref> [1805], A.C.21. </noteref>
</rootelement>
Can somebody help me please? Thanks in advance.

Answer 1

You are iterating over elemnts and anyone of them have square brackets:
When you make your first iteration, you get the element <year> and its value is 1902, not [1902]. Your square brackets are in the element <noteref>.
Change your XML to store the year as [1902] or change your for-each to print the square brackets manually.

Answer 2

Simply output the string value of the element to get all text, if you don't want the line breaks the also normalize the string:
<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="ref">
<noteref>
<xsl:value-of select="normalize-space()"/>
</noteref>
</xsl:template>
</xsl:transform>
Result is
<rootelement>
<pnote>
<P>
<noteref>[1902], A.C. 12</noteref>.</P>
</pnote>
<pnote>
<P>
<noteref>1902 B.A. 98</noteref>.</P>
</pnote>
<pnote>
<P> This is a text </P>
</pnote>
<pnote>
<P>
<noteref>[1912], 1 R.B. 160</noteref>, at pp. 165, 169.</P>
</pnote>
</rootelement>
Online at http://xsltransform.hikmatu.com/b4GWV2.