I'm looking for the simplest way to convert a string containing valid XML into an XmlElement object in C#.
How can you turn this into an XmlElement?
<item><name>wrench</name></item> 
5 Answers
Use this:
private static XmlElement GetElement(string xml) { XmlDocument doc = new XmlDocument(); doc.LoadXml(xml); return doc.DocumentElement; } Beware!! If you need to add this element to another document first you need to Import it using ImportNode.
3 Comments
Jimmy Hoffa
Wont this fail if there's no <?xml version bla bla> tag at the beginning? If he just has an xml fragment I don't think this will work..
dtb
@Jimmy Hoffa: IIRC LoadXml takes any well-formed XML fragment that contains exactly one XML element at the top level.
<?xml at the beginning is not required.
Suncat2000
To be clear, this works only if the XML fragment does not contain more than a single element, so it can be treated as a root element. Otherwise, it throws an XmlException stating, "There are multiple root elements." For example, loading "<item/><item/>" this way would fail.
Suppose you already had a XmlDocument with children nodes, And you need add more child element from string.
XmlDocument xmlDoc = new XmlDocument(); // Add some child nodes manipulation in earlier // .. // Add more child nodes to existing XmlDocument from xml string string strXml = @"<item><name>wrench</name></item> <item><name>screwdriver</name></item>"; XmlDocumentFragment xmlDocFragment = xmlDoc.CreateDocumentFragment(); xmlDocFragment.InnerXml = strXml; xmlDoc.SelectSingleNode("root").AppendChild(xmlDocFragment); The Result:
<root> <item><name>this is earlier manipulation</name> <item><name>wrench</name></item> <item><name>screwdriver</name> </root> 
1 Comment
user430788
This seems to be the canonically correct way according to Ms docs. I dont know why it isnt the preferred solution
Use XmlDocument.LoadXml:
XmlDocument doc = new XmlDocument(); doc.LoadXml("<item><name>wrench</name></item>"); XmlElement root = doc.DocumentElement; (Or in case you're talking about XElement, use XDocument.Parse:)
XDocument doc = XDocument.Parse("<item><name>wrench</name></item>"); XElement root = doc.Root; 
3 Comments
Jimmy Hoffa
He wanted the element, and for XElement he can just do XElement.Parse(xmlString), but you're giving him a document not element.
dtb
@Jimmy Hoffa: If you have a document, it's straightforward to get the root element, no?
Jimmy Hoffa
Sure, was just saying your answer could be tailored to the posters question a little more in case it's not as easy for him as it is for us..
You can use XmlDocument.LoadXml() to do this.
Here is a simple examle:
XmlDocument xmlDoc = new XmlDocument(); xmlDoc.LoadXml("YOUR XML STRING"); 
Comments
I tried with this snippet, Got the solution.
// Sample string in the XML format String s = "<Result> No Records found !<Result/>"; // Create the instance of XmlDocument XmlDocument doc = new XmlDocument(); // Loads the XML from the string doc.LoadXml(s); // Returns the XMLElement of the loaded XML String XmlElement xe = doc.DocumentElement; // Print the xe Console.out.println("Result :" + xe); If any other better/ efficient way to implement the same, please let us know.
Thanks & Cheers
XmlElement[]after the svcutil generates your proxy, you are kind of forced weirdness.