Hi,

Can anyone please help me?

I have the xml file:

<Issue>
     <IssueHistory_CreationDate>2004-01-01</IssueHistory_CreationDate>
      <Status>
           <Name>Find</Name>
                <Amount>28</Amount>
           <Name>resolved</Name>
                <Amount>0</Amount>
            <Name>Finished</Name>
                <Amount>0</Amount>
   </Status>
    <IssueHistory_CreationDate>2004-01-02</IssueHistory_CreationDate>
    <Status>
           <Name>Find</Name>
               <Amount>6</Amount>
          <Name>Resolved</Name>
                <Amount>6</Amount>
           <Name>Finished</Name>
                <Amount>4</Amount>
   </Status>

But I want to sort the Fined, Resolved and Finished amount
with xslt.

I want that the xslt file generate my xml file to this:

<string>Fined</string>
<Number>28</Number>
<Number>6</Number>

<string>solved</string>
<Number>0</Number>
<Number>6</Number>

<string>Finished</string>
<Number>0</Number>
<Number>4</Number>

Does anyone know how I can do this?

If the set of status values is fixed, just do

<xsl:template name="statusReport">
  <xsl:param name="status"/>
  <string><xsl:value-of select="$status"/></string>
  <xsl:for-each select="IssueHistory/Status[Name=$status]">
    <Number><xsl:value-of select="Amount"/></Number>
  </
</

<xsl:template match="Issue">
  <xsl:call-template name="statusReport">
    <xsl:with-param name="status">Find</xsl:with-param>
  </xsl:call-template>
  <xsl:call-template name="statusReport">
    <xsl:with-param name="status">Resolved</xsl:with-param>
  </xsl:call-template>

etc.

If the set of status values isn't fixed, then you have a grouping problem (or you could tackle it as a grouping problem anyway) - use xsl:for-each-group in XSLT 2.0, or see www.jenitennison.com/xslt/grouping

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

Hi Mhkay,

what do mean with If the set of status values isn't fixed?

I want to show al the 3, Fined, Solved and Finished. The file must show the Fined, Solved and Finished with grouping their amount.

If your input documents always use "Fined" "Solved" and "Finished" then that's what I call a fixed set of values. If different documents might use different values, for example "Complete" and "Postponed" then you need to use grouping techniques.

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

Hi again,

okee, thanks. But wich of the two should I use?

<xsl:template name="statusReport">
  <xsl:param name="status"/>
  <string><xsl:value-of select="$status"/></string>
  <xsl:for-each select="IssueHistory/Status[Name=$status]">
    <Number><xsl:value-of select="Amount"/></Number>
  </
</

<xsl:template match="Issue">
  <xsl:call-template name="statusReport">
    <xsl:with-param name="status">Find</xsl:with-param>
  </xsl:call-template>
  <xsl:call-template name="statusReport">
    <xsl:with-param name="status">Resolved</xsl:with-param>
  </xsl:call-template>

Eh?

I gave you two templates one of which calls the other. If you can't work out how to integrate those into a stylesheet, then you need to do some more basic reading before you do any more coding.

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

Hi again,

Yes, I am busy with basic reading.

first of all thank you for your help.

I have this:

<?xml version="1.0" encoding="UTF-8" ?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:output method="xml" indent="yes"/>

<xsl:template name="statusReport">
<xsl:param name="status"/>
<string><xsl:value-of select="$status"/></string>
<xsl:for-each select="IssueHistory/Status[Name=$status]">
<Number><xsl:value-of select="Amount"/></Number>

 <row >
        <xsl:for-each select="Issue/IssueHistory_CreationDate">
         <String><xsl:apply-templates/></String>
        </xsl:for-each>
   </row>

<xsl:template match="Issue">
<xsl:call-template name="statusReport">
<xsl:with-param name="status">Find</xsl:with-param>
</xsl:call-template>

<xsl:call-template name="statusReport">
<xsl:with-param name="status">Resolved</xsl:with-param>
</xsl:call-template>

</xsl:template>
</xsl:stylesheet>

Is this the correct xslt?

You need to add the missing end tags to make it well-formed. I used </ as a short-hand because that's what I use in an XML editor, and I assume people can fill in the gaps.

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

<?xml version="1.0" encoding="UTF-8" ?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:output method="xml" indent="yes"/>

<xsl:template name="statusReport">
<xsl:param name="status"/><string><xsl:value-of select="$status"/></string>

<xsl:for-each select="Issue/Status[Name=$status]"><Number><xsl:value-of select="Amount"/></Number>
</xsl:for-each>
<xsl:template match="Issue"></xsl:template>

<Chart_data>

 <row >
        <xsl:for-each select="Issue/IssueHistory_CreationDate">
         <String><xsl:apply-templates/></String>
        </xsl:for-each>
   </row>

<row>
<xsl:call-template name="statusReport">
<xsl:with-param name="status">Find</xsl:with-param>
</xsl:call-template>
</row>

<row>
<xsl:call-template name="statusReport">
<xsl:with-param name="status">Resolved</xsl:with-param>
</xsl:call-template>
</row>

</Chart_data>
</xsl:template>
</xsl:stylesheet>

When I translate this withe the xml file i got a error. It says:
Invalid at the top level of the document. Error processing resource 'file:///C:/Documents and Settings/Desktop/resu...

Try closing the first template before you open the second one.

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