Hi all,
Newbie to vb so please be gentle....
I am trying to transform some xml using xslcompliedtransform and the
xsltargumentlist to another xml file. The file is created, but with only a
root element
The main projects.xml file is loaded into a data grid view with some basic
details, once a row is clicked, i want to filter the main xml file to
create a html report containing all the xml data for the selected entry.
Obviously, not the full correct xml, but is an edited structure...
<?xml version="1.0" encoding="UTF-8"?><Root>
<Row A1="123456">
<F48>Open</F48>
<F49>Project One</F49>
</Row>
<Row A1="654321">
<F48>Open</F48>
<F49>Project One A</F49>
</Row>
<Row A1="223344">
<F48>Closed</F48>
<F49>Project Two</F49>
</Row></Root>
The XSL being used:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes" />
<xsl:decimal-format name="NN" NaN="0" />
<xsl:param name="ID" />
<xsl:template match="@* | node()">
<xsl:copy>
<xsl:apply-templates select="/@* | node()" />
</xsl:copy>
</xsl:template>
<xsl:template match="Root">
<xsl:copy>
<xsl:apply-templates select="Row[@A1=$ID]" />
</xsl:copy>
</xsl:template>
<xsl:template match="Row">
<Row>
<xsl:apply-templates select="@A1 | F49 | F48" />
</Row>
</xsl:template>
</xsl:stylesheet>
The vb code running on a button click is
Dim oXSL As New XslCompiledTransform()
oXSL.Load("findProject.xsl")Dim oXslArg As New XsltArgumentList()
oXslArg.AddParam("ID", "", dgList.Rows(e.RowIndex).Cells(0).Value
oXSL.Transform(New XPathDocument("projects.xml"),
XmlWriter.Create(dgList.Rows(e.RowIndex).Cells(0).Value & ".xml"))
The file that is created is
<?xml version="1.0" encoding="utf-8"?><Root />
When running the transformation in XML Notepad or other xml editor, the
transform works perfectly....any ideas people? Or any suggestions for a better
method?
Many thanks
Graham
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