Passando parâmetros para XSLT Stylesheet via .NET

Question

Eu estou tentando passar um parâmetro para uma folha de estilo XSLT, mas tudo que eu estou recebendo é um documento XML vazio quando o documento é transformado usando XslCompiledTransform.

Este é o método C # usado para adicionar os parâmetros (depois de adicionar em sugestões das pessoas)

private static void CreateHierarchy(string manID)
    {

        string man_ID = manID;

        XsltArgumentList argsList = new XsltArgumentList();
        argsList.AddParam("Boss_ID","",man_ID);

        XslCompiledTransform transform = new XslCompiledTransform();
        transform.Load("htransform.xslt");

        using (StreamWriter sw = new StreamWriter("output.xml"))
        {
            transform.Transform("LU AIB.xml", argsList, sw);
        } 


    }

e aqui é a folha de estilo. O parâmetro que eu estou passando é 'Boss_ID'

   <?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 match="OrgDoc">
     <xsl:param name="Boss_ID"></xsl:param>
    <xsl:processing-instruction name="xml-stylesheet">
    <xsl:text>type="text/xsl" href="..\styles\orgcharts.xsl" </xsl:text>
  </xsl:processing-instruction>
    <OrgDoc>
      <xsl:for-each select="PosDets[@OC_Man = $Boss_ID]">
      <PosDets OC_Pos="{@OC_Pos}" OC_Sub="{@OC_Sub}" OC_Man="{@OC_Man}" OC_Ttl="{@OC_Ttl}" OC_Rnk="{@OC_Rnk}" OC_Bdg="{@OC_Bdg}" OC_Fnd="{@OC_Fnd}"   OC_OL3="{@OC_OL3}"    OC_Tmp="{@OC_Tmp}">
          <xsl:apply-templates select="../PosDets">
            <xsl:with-param name="mgrid" select="@OC_Pos"/>
          </xsl:apply-templates>
        </PosDets>  
      </xsl:for-each>
    </OrgDoc>
  </xsl:template>
  <xsl:template match="PosDets" > 
    <xsl:param name="mgrid" />
    <xsl:if test="@OC_Man=$mgrid" >
      <PosDets OC_Pos="{@OC_Pos}" OC_Sub="{@OC_Sub}" OC_Man="{@OC_Man}" OC_Ttl="{@OC_Ttl}" OC_Rnk="{@OC_Rnk}" OC_Bdg="{@OC_Bdg}" OC_Fnd="{@OC_Fnd}"   OC_OL3="{@OC_OL3}"    OC_Tmp="{@OC_Tmp}">
        <xsl:apply-templates select="../PosDets">
          <xsl:with-param name="mgrid" select="@OC_Pos"/>
        </xsl:apply-templates>
      </PosDets>  
    </xsl:if>
  </xsl:template>


   </xsl:stylesheet>  

Eu não posso postar tudo do documento de entrada como é informação confidencial, mas aqui está uma breve versão higienizado

<OrgDoc><PosDets OC_Pos="161" OC_Man="9" OC_Ttl="Boss" OC_Rank="" OC_OL3="LU AIB" OC_Bdg="Has Budget" OC_Fnd="Payroll" OC_Sub="" OC_Tmp="" /><PosDets OC_Pos="190" OC_Man="161" OC_Ttl="Boss" OC_Rank="" OC_OL3="LU AIB" OC_Bdg="Has Budget" OC_Fnd="Payroll" OC_Sub="" OC_Tmp="" /><PosDets OC_Pos="199" OC_Man="190" OC_Ttl="Boss" OC_Rank="" OC_OL3="LU AIB" OC_Bdg="Has Budget" OC_Fnd="Payroll" OC_Sub="" OC_Tmp="" /></OrgDoc>

alguém pode ajudar?

Graças

Answer 1

You need to define the parameter within your XSLT and you also need to pass the XsltArgumentList as an argument to the Transform call:

private static void CreateHierarchy(string manID)
{
    string man_ID = manID;

    XsltArgumentList argsList = new XsltArgumentList();
    argsList.AddParam("Boss_ID", "", man_ID);

    XslCompiledTransform transform = new XslCompiledTransform(true);
    transform.Load("htransform.xslt");

    using (StreamWriter sw = new StreamWriter("output.xml"))
    {
        transform.Transform("LU AIB.xml", argsList, sw);
    }
}

Please note that the xsl:param must be defined below the xsl:stylesheet element:

<?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:param name="Boss_ID"></xsl:param>

  <xsl:template match="OrgDoc">

     <!-- template body goes here -->

  </xsl:template>


</xsl:stylesheet>

---

This simple XSLT sample will create just a small output document containing one XML node with its contents set to the value of your parameter. Have a try:

<?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:param name="Boss_ID"></xsl:param>

  <xsl:template match="/">
    <out>
      <xsl:value-of select="$Boss_ID" />
    </out>
  </xsl:template>

</xsl:stylesheet>

Answer 2

you probably need to define the param at the top of the XSLT:

<xsl:param name="Boss_ID" />
<OrgDoc>
 //rest of the XSLT
</OrgDoc>

See this link

http://projects.ischool.washington.edu/tabrooks/545/2004Autumn/ContentManagement/PassingParameters.htm

Not a great example but the best I could find with a quick google.