How to change XML node values

Question

I have an XML (this is exactly what it looks like):

<PolicyChangeSet schemaVersion="2.1" username="" description="">
    <Attachment name="" contentType="">
        <Description/>
        <Location></Location>
    </Attachment>
</PolicyChangeSet>

This is on the user's machine.

I need to add values to each node: username, description, attachment name, contenttype, and location.

This is what I have so far:

string newValue = string.Empty;
XmlDocument xmlDoc = new XmlDocument();

xmlDoc.Load(filePath);
XmlNode node = xmlDoc.SelectSingleNode("/PolicyChangeSet");
node.Attributes["username"].Value = AppVars.Username;
node.Attributes["description"].Value = "Adding new .tiff image.";
node.Attributes["name"].Value = "POLICY";
node.Attributes["contentType"].Value = "content Typeeee";

//node.Attributes["location"].InnerText = "zzz";

xmlDoc.Save(filePath);

Any help?

Jan · Accepted Answer · 2012-08-10 14:17:42Z

15

With XPath. XmlNode node = xmlDoc.SelectSingleNode("/PolicyChangeSet"); selects your root node.

answered Aug 10, 2012 at 14:17

Jan

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7 Comments

JJ. Over a year ago

that worked :) ... i can only accept your answer in 10 mins tho. thx Jan!

JJ. Over a year ago

how would I add a value for "location" though? it's just between the <> </> ... ??

Jan Over a year ago

Anytime :) Look at the InnerText property of XmlNode.

JJ. Over a year ago

by the way, username and description are working OK but im getting an error when i need to change the attachment NAME, i edited my post with my code, please take a look

Jan Over a year ago

Location is a node, not an attribute. Also XPath is case sensitive. node = xmlDoc.SelectSingleNode("/PolicyChangeSet/Attachment/Location"); and then node.InnerText = "myLocation" is the way to go here

|

Igor Popov · Accepted Answer · 2022-08-30 11:47:25Z

6

xmlDoc.SelectSingleNode("/PolicyChangeSet/Attachment/Description").InnerText = "My Description";
xmlDoc.SelectSingleNode("/PolicyChangeSet/Attachment/Location").InnerText = "My Location";

edited Aug 30, 2022 at 11:47

Igor Popov

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answered Nov 12, 2014 at 1:57

Ramzay

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Comments

JJ. · Accepted Answer · 2012-08-10 14:32:04Z

4

Got it with this -

xmlDoc.Load(filePath);
            XmlNode node = xmlDoc.SelectSingleNode("/PolicyChangeSet");
            node.Attributes["username"].Value = AppVars.Username;
            node.Attributes["description"].Value = "Adding new .tiff image.";

            node = xmlDoc.SelectSingleNode("/PolicyChangeSet/Attachment");
            node.Attributes["name"].Value = "POLICY";
            node.Attributes["contentType"].Value = "content Typeeee";
xmlDoc.Save(filePath);

answered Aug 10, 2012 at 14:32

JJ.

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Comments

Nickon · Accepted Answer · 2012-08-10 14:36:59Z

4

Use LINQ To XML:)

XDocument doc = XDocument.Load(path);
IEnumerable<XElement> policyChangeSetCollection = doc.Elements("PolicyChangeSet");

foreach(XElement node in policyChangeSetCollection)
{
   node.Attribute("username").SetValue(someVal1);
   node.Attribute("description").SetValue(someVal2);
   XElement attachment = node.Element("attachment");
   attachment.Attribute("name").SetValue(someVal3);
   attachment.Attribute("contentType").SetValue(someVal4);
}

doc.Save(path);

answered Aug 10, 2012 at 14:36

Nickon

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Comments

Robert Zahm · Accepted Answer · 2012-08-10 14:21:25Z

1

In your SelectSingleNode method, you need to provide an XPath expression the find the node that you are looking to select. If you Google XPath you will find many resources for this.

http://www.csharp-examples.net/xml-nodes-by-name/

If you need to add these to each node, you can start at the top and iterate over all of the children.

http://msdn.microsoft.com/en-us/library/system.xml.xmlnode.aspx

answered Aug 10, 2012 at 14:21

Robert Zahm

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Comments

Mohan Krishna Eluri · Accepted Answer · 2015-08-13 10:34:10Z

0

For setting value to XmlNode: 
 XmlNode node = xmlDoc.SelectSingleNode("/PolicyChangeSet");
            node["username"].InnerText = AppVars.Username;
            node["description"].InnerText = "Adding new .tiff image.";
            node["name"].InnerText = "POLICY";
            node["contentType"].InnerText = "content Typeeee";

For Getting value to XmlNode: 
username=node["username"].InnerText

answered Aug 13, 2015 at 10:34

Mohan Krishna Eluri

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