Generating a sequence that will sort in alphabetic order - xslt-2.0

Question

Is there a built-in or simple way to generate a sequence of values that will sort in alphabetical order?
As an example, the following item elements have a child sequence element intended for sorting on. The data is being generated for another system that requires this information, since it does not work with the document ordering.
<list>
<item>
<sequence>1</sequence>
</item>
<item>
<sequence>2</sequence>
</item>
<item>
<sequence>3</sequence>
</item>
<item>
<sequence>4</sequence>
</item>
...
<item>
<sequence>14</sequence>
</item>
<item>
<sequence>15</sequence>
</item>
Instead of sorting numerically, the system sorts alphabetically and it cannot change this behavior. As a result the items are sorted as 1,10,11,12,13,14,15,2,3,4,5,6,7,8,9
Furthermore, the actual number of items is unbounded, so there needs to be general way of generating an alphabetic sequence that will sort in the right order. That said, the number of items should be relatively small, under 1000.
This is what I have so far:
<xsl:variable name="idchars" as="xs:string *"
select="
('0','1','2','3','4','5','6','7','8','9',
'A','B','C','D','E','F','G','H','I','J','K','L','M',
'N','O','P','Q','R','S','T','U','V','W','X','Y','Z',
'a','b','c','d','e','f','g','h','i','j','k','l','m',
'n','o','p','q','r','s','t','u','v','w','x','y','z')
"
/>
<xsl:variable name="idlen" as="xs:integer" select="count($idchars)" />
<xsl:function name="f:genId" as="xs:string">
<xsl:param name="pSeq" as="xs:integer" />
<xsl:variable name="vLen" as="xs:integer" select="$pSeq idiv $idlen" />
<xsl:value-of>
<xsl:for-each select="1 to $vLen">
<xsl:sequence select="$idchars[$idlen]" />
</xsl:for-each>
<xsl:sequence select="$idchars[$pSeq mod $idlen]" />
</xsl:value-of>
</xsl:function>

Answer 1

You could use format-number() in order to generate a left-padded sequence of numbers with a variable length of leading zeros, then convert each of those digits into a letter using codepoints-to-string(), and then join them together with string-join():
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:f="local">
<xsl:output indent="yes"/>
<xsl:function name="f:genId" as="xs:string">
<xsl:param name="pSeq" as="xs:integer" />
<xsl:param name="digits" as="xs:integer"/>
<xsl:variable name="picture" select="string-join( ((1 to $digits) ! '0'), '')" />
<xsl:variable name="padded-string" select="format-number($pSeq, $picture)" as="xs:string"/>
<xsl:variable name="padded-letter-seq" select="(1 to string-length($picture)) ! substring($padded-string, ., 1) ! codepoints-to-string(xs:integer(.) + 65)" as="xs:string*"/>
<xsl:sequence select="string-join($padded-letter-seq, '')"/>
</xsl:function>
<xsl:function name="f:genId" as="xs:string">
<xsl:param name="pSeq" as="xs:integer"/>
<xsl:sequence select="f:genId($pSeq, 4)"/>
</xsl:function>
<xsl:template match="/">
<xsl:sequence select="f:genId(3)"/>
<xsl:sequence select="f:genId(3,5)"/>
</xsl:template>
</xsl:stylesheet>

Answer 2

Just use the <xsl:sort> xslt element, simply like this:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/*">
<xsl:copy>
<xsl:apply-templates select="*">
<xsl:sort select="sequence"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
<xsl:template match="node()|#*">
<xsl:copy>
<xsl:apply-templates select="node()|#*"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
When this transformation is applied on the provided XML document:
<list>
<item>
<sequence>1</sequence>
</item>
<item>
<sequence>2</sequence>
</item>
<item>
<sequence>3</sequence>
</item>
<item>
<sequence>4</sequence>
</item>
...
<item>
<sequence>14</sequence>
</item>
<item>
<sequence>15</sequence>
</item>
</list>
the wanted, correct result is produced:
<list>
<item>
<sequence>1</sequence>
</item>
<item>
<sequence>14</sequence>
</item>
<item>
<sequence>15</sequence>
</item>
<item>
<sequence>2</sequence>
</item>
<item>
<sequence>3</sequence>
</item>
<item>
<sequence>4</sequence>
</item>
</list>
Do note:
This:
<xsl:sort select="sequence"/>
is equivalent to:
<xsl:sort select="sequence" data-type="text"/>
as "text" is the default value for the data-type attribute.
If it is necessary to sort numerically, one should explicitly use:
<xsl:sort select="sequence" data-type="number"/>