Hi Michael,

Thanks, Yeaaah it is working.

Thanks a lot. Youre great.

Hi michael,

do you also know how I translate 12-01-2004T12:00:00 to only 12-01-2004 withoudt te time in xslt?

<Issue> <IssueHistory_CreationDate>2004-01-01T12:00:00</IssueHistory_CreationDate>
      <Status>
      <Name>test</Name>
   </Status> <IssueHistory_CreationDate>2004-01-02T12:00:00</IssueHistory_CreationDate>
    <Status>
        <Name>test</Name>
   </Status>
   </Issue>

thanks.

Use the substring-before() function.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

Yes i know that. But I want select all the IssueHistory_Creationdate and not one. And also it must be iteratieve. I dont't want to mention the date in the variable.

It must be iteratieve that alle the dates is been showing.


<xsl:variable name = "A" >2004-01-01T</xsl:variable>
 <xsl:for-each select="Issue/IssueHistory_CreationDate">
                <String><xsl:apply-templates/></String>
                <xsl:value-of select = "substring-before($A, 'T')"/>
  </xsl:for-each>

this is the result:

<?xml version="1.0" encoding="UTF-8"?>
<String>2004-01-01T:00:00</String>2004-01-01
<String>2004-01-02T:00:00</String>2004-01-01

I want to show this:
<String>2004-01-01</String>
<String>2004-01-02</String>

I don't really understand your question. It looks to me as if you want

 <xsl:for-each select="Issue/IssueHistory_CreationDate">
                <String><xsl:value-of select = "substring-before(., 'T')"/></String>

  </xsl:for-each>

But you make it seem a lot more complicated than that.


Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

Hi Michael,

Thank you so very much. It works.