Question

I have some short form day names like so:

M -> Monday
T -> Tuesday
W -> Wednesday
R -> Thursday
F -> Friday
S -> Saturday
U -> Sunday

How can I convert an xml element like <days>MRF</days> into the long version <long-days>Monday,Thursday,Friday</long-days> using xslt?

Update from comments

Days will not be repeated

Answer 1

This stylesheet

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
 xmlns:d="day"
 exclude-result-prefixes="d">
    <d:d l="M" n="Monday"/>
    <d:d l="T" n="Tuesday"/>
    <d:d l="W" n="Wednesday"/>
    <d:d l="R" n="Thursday"/>
    <d:d l="F" n="Friday"/>
    <d:d l="S" n="Saturday"/>
    <d:d l="U" n="Sunday"/>
    <xsl:variable name="vDays" select="document('')/*/d:d"/>
    <xsl:template match="days">
        <long-days>
            <xsl:apply-templates
                 select="$vDays[contains(current(),@l)]"/>
        </long-days>
    </xsl:template>
    <xsl:template match="d:d">
        <xsl:value-of select="@n"/>
        <xsl:if test="position()!=last()">,</xsl:if>
    </xsl:template>
</xsl:stylesheet>

With this input:

<days>MRF</days>

Output:

<long-days>Monday,Thursday,Friday</long-days>

Edit: For those who wander, retaining the sequence order:

<xsl:variable name="vCurrent" select="current()"/>
<xsl:apply-templates
         select="$vDays[contains($vCurrent,@l)]">
    <xsl:sort select="substring-before($vCurrent,@l)"/>
</xsl:apply-templates>

Note: Because days wouldn’t be repeated, this is the same as looking up for item existence in sequence with empty string separator.