recursiv search in a xml file

Posted on 2013-01-19
Last Modified: 2013-02-08
Hi Everybody,

i'm looking for a solution to read an xml file recursiv.  (see attached xml file).
What i have is the "Filename", Filename = 193478618937468178.000.
Beginning from the the "Filename" i need the ID of the next node on the top.
From this ID i want to go to the next Childnode on the top, and so on.

Is there any idea how could i do this with Xpath and C# .net?

Filename = 193478618937468178.000 --> goto ID = 2657
ID = 2657 --> goto ID = 1392
ID = 1392 --> goto ID = 8345
ID = 8345 --> goto ID = 1242
ID = 1242 --> goto ID = 39

Thank you in advance
Regards Bounty
Question by:bounty457
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LVL 60

Expert Comment

by:Geert Bormans
ID: 38796780
The most efficient way to do this, I think is by running a simple XSLT from C# on the document
You can pass in the file name as a parameter

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl=""
    <xsl:output method="text"/>
    <xsl:strip-space elements="*"/>
    <xsl:param name="fname">193478618937468178.000</xsl:param>
    <xsl:key name="object" match="Object" use="Fields/Field[@name='Filename']"/>
    <xsl:template match="/">
        <xsl:variable name="root-node" select="key('object', $fname)"/>
        <xsl:text>Filename = </xsl:text>
        <xsl:value-of select="$fname"/>
        <xsl:text> --> goto ID = </xsl:text>
        <xsl:apply-templates select="$root-node"/>       
    <xsl:template match="*[@id]">
        <xsl:value-of select="@id"/>
        <xsl:if test="ancestor::*[@id]">
            <xsl:text>&#10;From&#10;ID = </xsl:text>
            <xsl:value-of select="@id"/>
            <xsl:text> --> goto ID = </xsl:text>
            <xsl:apply-templates select="ancestor::*[@id][1]"></xsl:apply-templates>

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Author Comment

ID: 38796829
Thank you for the answer.

Could you please short explain how must i use the code from C#?
i want do write the data for each "filename" a one datatable after parsing the xml.

Regards Bountry
LVL 60

Accepted Solution

Geert Bormans earned 400 total points
ID: 38796874
Here is an example

put the name of your XSLT file there

drop the two DateTime lines

      argList.AddParam("discount", "", discountDate.ToString());
should be
      argList.AddParam("fname", "", "193478618937468178.000");
you can put in there any value you want

 XmlWriter writer = XmlWriter.Create("orderOut.xml");
is where you want the result to be

xslt.Transform(new XPathDocument("order.xml"), argList, writer);
pull in the right file
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Author Comment

ID: 38797985

the sample from MSDN work right. If i change the .xsl and .xml File
i get the message:
"Leerzeichen können aus bereits geladenen Eingabedokumenten nicht entfernt werden. Stellen Sie das Eingabedokument stattdessen als XmlReader bereit."
"Use XMLReader - Empty Spaces can't delete from the source file"

I looked at MSDN:
I use the  XML Reader with the following code:
// Create the XslCompiledTransform and load the style sheet.
      XslCompiledTransform xslt = new XslCompiledTransform();
      string stylesheet = "Variant3.xsl";

//Create the reader to load the stylesheet.  
//Move the reader to the xsl:stylesheet node.
          XmlTextReader reader = new XmlTextReader(stylesheet);
// Create the XsltArgumentList.
          XsltArgumentList argList = new XsltArgumentList();
// Add my first Filename to the Arglist
          argList.AddParam("fname", "", "193478618937468178.000");
// Create an XmlWriter to write the output.             
         XmlWriter writer = XmlWriter.Create("Variant3_Out.xml");
// Transform the file.
         xslt.Transform(new XPathDocument("xpath_Variant3.xml"), argList, writer); 

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Now i get the Message:

Das Stylesheet muss entweder mit einem 'xsl:stylesheet'- oder einem 'xsl:transform'-Element oder aber mit einem Literalergebniselement beginnen, welches das 'xsl:version'-Attribut aufweist. Dabei gibt das Präfix 'xsl' den Namespace '' an.
"The Stylesheet must start with: xsl:stylesheet or xsl:transform element or an literal with 'xsl:version'-Attribut. The präfix must have the namespace'
What must i add to the XSL File?

Regards Bounty
LVL 60

Expert Comment

by:Geert Bormans
ID: 38798010
apparently you have not loaded the XSLT correctly, since the above code I posted has all it needs
<xsl:stylesheet xmlns:xsl=""

Can you check that it finds the XSLT file?

Author Comment

ID: 38798033
Yes the File is found. The path is correct. It is a load Exception of the .xls File. It is working on your code?
I change the Variant3.xsl how you wrote.
I attached the screenshot from the error message.

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