Question

How do I convert a time string like

20101115083000 +0200

to

2010-11-15 08:30:00 +0200

using XSLT?

Answer 1

We use templates:

<xsl:template name="format-date-time">
    <xsl:param name="date" select="'%Y-%m-%dT%H:%M:%S%z'"/>
    <xsl:value-of select="substring($date, 9, 2)"/>
    <xsl:text>/</xsl:text>
    <xsl:value-of select="substring($date, 6, 2)"/>
    <xsl:text>/</xsl:text>
    <xsl:value-of select="substring($date, 1, 4)"/>
    <xsl:text> </xsl:text>
    <xsl:value-of select="substring($date, 12, 2)"/>
    <xsl:text>:</xsl:text>
    <xsl:value-of select="substring($date, 15, 2)"/>
    <xsl:text>:</xsl:text>
    <xsl:value-of select="substring($date, 18, 2)"/>
</xsl:template>

We call these templates like this:

<xsl:call-template name="format-date-time">
    <xsl:with-param name="date" select="./Startdate"/>
</xsl:call-template>

./Startdate is an XML date, but with the substring technique, I think you could solve your problem too.