Getting an xml attribute with XPAth using c#

Question

I'm trying to get an attribute from the XML but I'm getting an error:

Expression must evaluate to a node-test

I'm new to XPath.

XML file example:

<Message>
   <Head>
      <Sender Identifier="id" ref=""/>
   </Head>
</Message>

Code:

XPathDocument xDoc = new XPathDocument("documentName.xml");
XPathNavigator nav = xDoc.CreateNavigator();
XPathNodeIterator xpathNode = nav.Select("/Message/Head/Sender/[@Identifier]");
//This is where I get the error.

And I'm also not sure how to display the attribute value.

Welcome to StackOverflow: if you post code, XML or data samples, please highlight those lines in the text editor and click on the "code samples" button ( { } ) on the editor toolbar to nicely format and syntax highlight it! — marc_s
– marc_s, Commented Nov 29, 2012 at 8:48

Anirudha · Accepted Answer · 2012-11-29 08:53:46Z

2

It should be

/Message/Head/Sender[@Identifier]
                    ^
                    |->no need of / here since Identifier is an attribute of sender

So,

/Message/Head/Sender[@Identifier] would only select Sender if it has an Identifier attribute

/Message/Head/Sender[@Identifier='id'] would only select Sender if it has an Identifier attribute with value id

edited Nov 29, 2012 at 8:53

answered Nov 29, 2012 at 8:48

Anirudha

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1 Comment

Alexei Levenkov Over a year ago

+1. Good explanation. Usual note - if id contains single qoute you'll need to do concat tricks.

Murali N · Accepted Answer · 2012-11-29 08:46:07Z

1

Here is the wonderful tool to evaluate Xpath of the given XML

go to this link http://www.xmltoolbox.com/ and try to paste your xml in the textarea and move your cursor to the node you want Xpath.Xpath will be generated below

answered Nov 29, 2012 at 8:46

Murali N

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Comments

Rahul Rajput · Accepted Answer · 2012-11-29 08:47:16Z

-1

 XmlIO xml = new XmlIO(xmlPath);
                ProductName = xml.SelectNodeValue(@"//PRODUCTINFO/Application/@ProductName");
                IDictionary<string, string> keyValList = xml.SelectNodesList(@"//PRODUCTINFO/Application/Parameters");

Take a look

answered Nov 29, 2012 at 8:47

Rahul Rajput

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