Transform 20050416 to 2005-04-16

[J-M Andersson]

How can I transform.
<date>20050416</date>
to
<date>2005-04-16</date>
Today I do like this (but I guess there should be some better ways)

<xslt:template match="date">
<xslt:copy>
<xslt:value-of select="substring(.,1,4)"/>-
<xslt:value-of select="substring(.,5,2)"/>-
<xslt:value-of select="substring(.,7)"/>
</xslt:copy>
</xslt:template>

[Keith Davies]

J-M Andersson <no****@nospam.nospam> wrote:

How can I transform.
<date>20050416</date>
to
<date>2005-04-16</date>
Today I do like this (but I guess there should be some better ways)

<xslt:template match="date">
<xslt:copy>
<xslt:value-of select="substring(.,1,4)"/>-
<xslt:value-of select="substring(.,5,2)"/>-
<xslt:value-of select="substring(.,7)"/>
</xslt:copy>
</xslt:template>

<xsl:template match='date'>
<xsl:copy>
<xsl:value-of select='concat(substring(.,1,4),"-",
substring(.,5,2),"-",
substring(.,7,2))' />
</xsl:copy>
</xsl:template>

Not a *lot* better, mind. A little. XSLT is limited in its string-
processing capabilities.
Keith
--
Keith Davies "Trying to sway him from his current kook-
ke**********@kjdavies.org rant with facts is like trying to create
ke**********@gmail.com a vacuum in a room by pushing the air
http://www.kjdavies.org/ out with your hands." -- Matt Frisch

[J-M Andersson]

Keith Davies wrote:

<xsl:template match='date'>
<xsl:copy>
<xsl:value-of select='concat(substring(.,1,4),"-",
substring(.,5,2),"-",
substring(.,7,2))' />
</xsl:copy>
</xsl:template>

Not a *lot* better, mind. A little. XSLT is limited in its string-
processing capabilities.

Looks good to me, thanks for the hint.

/JM