Passing xml as parameter to xslt and replacing the xml [duplicate] - xslt

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Closed 10 years ago.
Possible Duplicate:
XSLT for replace attribute in XML element if it exists in another XML?
I have two XML:
First XML:
<FirstXML>
<catalog>
<cd>
<title>Empire Burlesque</title>
<artist>Bob Dylan</artist>
<country>USA</country>
<company>Columbia</company>
<price>10.90</price>
<year>1985</year>
</cd>
</catalog>
</FirstXML>
Second XML:
<SecondXML>
<catalog>
<cd>
<title>Hide your heart</title>
<artist>Bonnie Tyler</artist>
<country>UK</country>
<company>CBS Records</company>
<price>9.90</price>
<year>1988</year>
</cd>
</catalog>
</SecondXML>
I want to Transform my First XML using XSLT. The cd node of first xml should be replaced with cd node of second xml.
Resulting XML from XSLT transformation:
<FirstXML>
<catalog>
<cd>
<title>Hide your heart</title>
<artist>Bonnie Tyler</artist>
<country>UK</country>
<company>CBS Records</company>
<price>9.90</price>
<year>1988</year>
</cd>
</catalog>
</FirstXML>
Please help me with XSLT for this. I think we would need to pass the Second XML as a parameter to the XSLT, I am trying to do that. I am very new to XSLT so might not be coding it correctly.
Any inputs on how we can do this would be helpful. Thanks in Advance.

Answer 1

I think we would need to pass the Second XML as a parameter to the
XSLT.
This is possible, but not necessary at all.
A more convenient way is to pass to the transformation just the filepath to the document and use the XSLT document() function for parsing and accessing it:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:param name="pDocLink" select=
"'file:///c:/temp/delete/SecondXml.xml'"/>
<xsl:template match="/*">
<xsl:copy>
<xsl:copy-of select="document($pDocLink)/*/*[1]"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
When this transformation is applied on the provided "first" document:
<FirstXML>
<catalog>
<cd>
<title>Empire Burlesque</title>
<artist>Bob Dylan</artist>
<country>USA</country>
<company>Columbia</company>
<price>10.90</price>
<year>1985</year>
</cd>
</catalog>
</FirstXML>
and the "second" document is placed at: c:\temp\delete\SecondXml.xml :
<SecondXML>
<catalog>
<cd>
<title>Hide your heart</title>
<artist>Bonnie Tyler</artist>
<country>UK</country>
<company>CBS Records</company>
<price>9.90</price>
<year>1988</year>
</cd>
</catalog>
</SecondXML>
the wanted, correct result is produced:
<FirstXML>
<catalog>
<cd>
<title>Hide your heart</title>
<artist>Bonnie Tyler</artist>
<country>UK</country>
<company>CBS Records</company>
<price>9.90</price>
<year>1988</year>
</cd>
</catalog>
</FirstXML>

Answer 2

There is a document function, use it to work with several input sources.
The description of your problem is too vague to make more detailed recommendations. Try to search some tutorials, e.g. http://www.ibm.com/developerworks/xml/library/x-tipcombxslt/

Answer 3

You should be looking at the document() function. Passing an XML tree in as a parameter works is some XML systems, not others.
If you have a variable such as
<xsl:variable name="source" select="document('filename.xml')//catalog"/>
then accessing the variable $source would give you the root of your CD branch
EDIT
Just for clarity of my answer to Dimitre Novatchev, and as I said you could do it, although in 90% of times you would want to use the document function, it is possible to pass a node fragment in through to a stylesheet from a C# application with AddParam, and could be used when you are generating that data from your own application, and don't need to write it (or where it isn't practical).
So for a stylesheet such as:
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl">
<xsl:output method="xml" indent="yes"/>
<xsl:param name="bonnie_tyler"/>
<xsl:template match="#* | node()">
<xsl:copy>
<xsl:apply-templates select="#* | node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="cd[ancestor::FirstXML]">
<xsl:copy>
<xsl:apply-templates select="msxsl:node-set($bonnie_tyler)//cd/*"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
the First and Second XML as per your original question, you could use the pass the nodes in such as:
try
{
XslCompiledTransform xt = new XslCompiledTransform();
Assembly ass = Assembly.GetEntryAssembly();
Stream strm = ass.GetManifestResourceStream("ConsoleApplication1.testparam.xslt");
XmlTextReader xmr = new XmlTextReader(strm);
xt.Load(xmr);
xmr.Close();
XmlDocument originalDocument = new XmlDocument();
strm = ass.GetManifestResourceStream("ConsoleApplication1.FirstXML.xml");
originalDocument.Load(strm);
XmlDocument replacementNode = new XmlDocument();
strm = ass.GetManifestResourceStream("ConsoleApplication1.SecondXML.xml");
replacementNode.Load(strm);
XsltArgumentList arg = new XsltArgumentList();
arg.AddParam("bonnie_tyler", "", replacementNode);
StringBuilder htmlOutput = new StringBuilder();
XmlWriter writer = XmlWriter.Create(htmlOutput);
xt.Transform(originalDocument, arg, writer);
Console.Out.Write(htmlOutput.ToString());
}
which would give you your desired output (although obviously you wouldn't be loading the files from disk as here)