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XPath and XSD

Posted on 2006-11-01
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2,036 Views
Last Modified: 2013-11-19
I have an XML file in a treeview and I can get the XPATH of the node for every treenode. I want to traverse my XSD (on which this XML is based on) and show all the attributes this node has according to the XSD.

How can I navigate to the element in the XSD from the XPATH? XSD has all the elements as:

<xs:Element name="whatever">

Can XPath get to the "names"?
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Question by:srinivas_vemla
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5 Comments
 

Author Comment

by:srinivas_vemla
ID: 17852639
Any ideas pls...
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LVL 29

Expert Comment

by:anarki_jimbel
ID: 17853062
I believe  it can :)
If you want to select ALL "names" you use something like this:

//Element[@name]

'@' points to attributes.

See http://www.w3schools.com/xpath/xpath_syntax.asp


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Author Comment

by:srinivas_vemla
ID: 17853263
Are we talking of using the XpathNavigator or something?... what would a sample C# code be?
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Accepted Solution

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anarki_jimbel earned 300 total points
ID: 17853882
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Assisted Solution

by:joechina
joechina earned 200 total points
ID: 17858579
Here is what I have, not sure if it is what you want:

XPathDocument x = new XPathDocument("test.xsd");
XPathNavigator nav = x.CreateNavigator();
XmlNamespaceManager mngr = new XmlNamespaceManager(nav.NameTable);
mngr.AddNamespace("xs", "http://www.w3.org/2001/XMLSchema");
XPathNavigator resultNav = nav.SelectSingleNode("//xs:element",mngr);
string result = resultNav.GetAttribute("name", "");

result is MyElement

test.xsd is

<?xml version="1.0" encoding="utf-8"?>
<xs:schema id="test" targetNamespace="http://tempuri.org/test.xsd" elementFormDefault="qualified" xmlns="http://tempuri.org/test.xsd" xmlns:mstns="http://tempuri.org/test.xsd" xmlns:xs="http://www.w3.org/2001/XMLSchema">
  <xs:element name="MyElement">
    <xs:complexType>
      <xs:sequence />
    </xs:complexType>
  </xs:element>
</xs:schema>
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