Question

My problem is that I receive for example a date in a concatenated format:

Ex: 20050728 

And I have to retrieve it in a readable format through my xslt.

Ex. 28 July 2005 

I also have a similar question regards time.

Ex: 0004 

To be displayed as 00:04

How is this done?

Answer 1

Use the XPath substring() function as shown in the solution below:

<xsl:stylesheet version='1.0'  xmlns:xsl='http://www.w3.org/1999/XSL/Transform'>  <xsl:output method='text'/>  <!--                                           -->  <xsl:variable name='vrMonths'>    <m>January</m>    <m>February</m>    <m>March</m>    <m>April</m>    <m>May</m>    <m>June</m>    <m>July</m>    <m>August</m>    <m>September</m>    <m>October</m>    <m>November</m>    <m>December</m>  </xsl:variable>  <!--                                           -->  <xsl:variable name='vMonths' select=   'document('')/*/xsl:variable[@name='vrMonths']/*'/> <!--                                           -->     <xsl:template match='date'>       <xsl:value-of select=        'concat(substring(.,7), ' ',              $vMonths[number(substring(current(),5,2))], ' ',              substring(.,1,4))'        />     </xsl:template>  <!--                                           -->     <xsl:template match='time'>      <xsl:value-of select=      'concat(substring(.,1,2),':',substring(.,3))'/>     </xsl:template> </xsl:stylesheet> 

When the above transformation is applied on this XML document:

<t>  <date>20050728</date>  <time>0004</time> </t> 

the wanted result is produced:

 28 July 2005  00:04