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Parsing XML from youtube API in C#

I've parsed some of the XML that youtube API returns. The XML I'm trying to parse can be found on a URL like this http://gdata.youtube.com/feeds/api/videos?v=2&q=stackoverflow
The nodes I'm trying to get are yt:accessControl and I'm not sure how to do this. I've tried with SelectNodes but got an empty result.

For youtube, I think you don't need that - google has API wrapper specifically for .NET: https://developers.google.com/youtube/2.0/developers_guide_dotnet
But if you still wish to do it manually, LINQ to XML methods like Descendants and Element take an XName as an argument. There is a conversion from string to XName that is happening automatically for you. You can fix this by adding an XNamespace before the strings in your Descendants and Element calls. Watch out because you have 2 different namespaces at work.
XNamespace ns = "http://gdata.youtube.com/schemas/2007";
var xElements = XElement.Parse(File.ReadAllText(#"c:\test\youtube.xml"))
.Descendants(ns + "accessControl");
Without Linq it would look something like this:
var xdoc = new XmlDocument();
xdoc.Load(#"c:\test\youtube.xml");
var namespaceManager = new XmlNamespaceManager(xdoc.NameTable);
namespaceManager.AddNamespace("yt", "http://gdata.youtube.com/schemas/2007");
var xmlNodeList = xdoc.SelectNodes("//yt:accessControl", namespaceManager);

How to get element content using only XPath and C# .NET

I've found a lot of articles about how to get node content by using simple XPath expression and C#, for example:
XPath:
/bookstore/author/first-name
C#:
string xpathExpression = "/bookstore/author/first-name";
nodes = navigator.Select(xpathExpression);
I wonder how to get content that is inside of an element, and the same element is inside another element and another and another.
Just take a look on below code:
<Cell>
<CellContent>
<Para>
<ParaLine>
<String>ABCabcABC abcABC abc ABCABCABC.</string>
</ParaLine>
</Para>
</CellContent>
</Cell>
I only want to extract content ABCabcABC abcABC abc ABCABCABC. from String element.
Do you know how to resolve problem by use XPath expression and .Net C#?

After googling c# .net xpath for few seconds you'll find this article, which provides example which you can easily modify to use XPathDocument, XPathNavigator and XPathNavigator::SelectSingleNode():
XPathNavigator nav;
XPathDocument docNav;
string xPath;
docNav = new XPathDocument("c:\\books.xml");
nav = docNav.CreateNavigator();
xPath = "/Cell/CellContent/Para/ParaLine/String/text()";
string value = nav.SelectSingleNode(xPath).Value
I recommend more reading on xPath syntax. Much more.

navigator.SelectSingleNode("/Cell/CellContent/Para/ParaLine/String/text()").Value

You can use Linq to XML as well to get value of specified element
var list = XDocument.Parse("xml string").Descendants("ParaLine")
.Select(x => x.Element("string").Value).ToList();
From above query you will get value of all the string element which are inside ParaLine tag.

XPath problem, getting "expression must evaluate to a node-set." error

I'm having trouble retrieving a single node by its explicit XPath that I have already found by other ways. I have node and I can get its XPath, but when I try to retrieve that same node again this time via node.XPath it gives the "expression must evaluate to a node-set" error. Shouldn't this work? I'm using HtmlAgilityPack in C# btw for the HtmlDocument.
HtmlDocument doc = new HtmlDocument();
doc.Load(#"..\..\test1.htm");
HtmlNode node = doc.DocumentNode.SelectSingleNode("(//node()[#id='something')])[first()]");
HtmlNode same = doc.DocumentNode.SelectSingleNode(node.XPath);
BTW: this is the value of node.XPath:
"/html[1]/body[1]/table[1]/tr[1]/td[1]/div[1]/div[1]/div[2]/table[1]/tr[1]/td[1]/div[1]/div[1]/table[1]/tr[1]/td[1]/div[1]/div[1]/div[4]/div[2]/div[1]/div[1]/div[4]/#text[2]"

I was able to get it working by replacing #text with the function text(). I'm not sure why it didn't just emit the XPath that way in the first place.
HtmlNode same = doc.DocumentNode.SelectSingleNode(node.XPath.Replace("#text","text()");

Your XPath ends in "#text[2]", which means "the second 'text' attribute". Attributes aren't nodes, they're node metadata.
This is a common problem I've had with XPath: wanting the value of an attribute while the XPath operation absolutely has to extract a node.
The solution I've used for this is to wrap my XPath fetching with something that detects and strips off the attribute portion of the string (via a myXPathString.LastIndexOf( "#" ) method call) and then uses the truncated myXPathString to fetch the node and collect the desired attribute value as a second step.
Hope that helps,
J

Get XmlNode Open Tag with Attributes

Is it possible to get the open tag from a XmlNode with all attributes, namespace, etc?
eg.
<root xmlns="urn:..." rattr="a">
<child attr="1">test</child>
</root>
I would like to retrieve the entire opening tag, exactly as retrieved from the original XML document if possible, from the XmlNode and later the closing tag. Both as strings.
Basically XmlNode.OuterXml without the child nodes.
EDIT
To elaborate, XmlNode.OuterXml on a node that was created with the XML above would return the entire XML fragment, including child nodes as a single string.
XmlNode.InnerXml on that same fragment would return the child nodes but not the parent node, again as a single string.
But I need the opening tag for the XML fragment without the children nodes. And without building it using the XmlAttribute array, LocalName, Namespace, etc.
This is C# 3.5
Thanks

Is there some reason you can't simply say:
string s = n.OuterXml.Substring(0, n.OuterXml.IndexOf(">") + 1);

I think the simplest way would be to call XmlNode.CloneNode(false) which (according to the docs) will clone all the attributes but not child nodes. You can then use OuterXml - although that will give you the closing tag as well.
For example:
using System;
using System.Xml;
public class Test
{
static void Main()
{
XmlDocument doc = new XmlDocument();
doc.LoadXml(#"<root xmlns='urn:xyz' rattr='a'>
<child attr='1'>test</child></root>");
XmlElement root = doc.DocumentElement;
XmlNode clone = root.CloneNode(false);
Console.WriteLine(clone.OuterXml);
}
}
Output:
<root xmlns="urn:xyz" rattr="a"></root>
Note that this may not be exactly as per the original XML document, in terms of the ordering of attributes etc. However, it will at least be equivalent.

How about:
xmlNode.OuterXML.Replace(xmlNode.InnerXML, String.Empty);
Poor man's solution :)

Select XMLNode by Id using XPath

I have the below fragement of XML, notice that the Reference node holds a URI which links to the Id attribute of the Body node.
<Reference URI="#Body">
<SOAP-ENV:Body Id="Body" xmlns:SOAP-ENV="http://www.dingo.org">
<ns0:Add xmlns:ns0="http://www.moo.com">
<ns0:a>2</ns0:a>
<ns0:b>3</ns0:b>
</ns0:Add>
</SOAP-ENV:Body>
If I had the value of the URI attribute how would I then get the whole Body XMLNode? I presume this would be best done via an XPath epression but haven't any clue on XPath. Note that the XML will not always be so simple. I'm doing this in c# btw :)
Any ideas?
Thanks
Jon
EDIT: I wouldn't know the XML structure or namespaces before hand, all I would know is that the reference element has the ID of the xmlNode i want to retrieve, hope this is sligtly clearer.

You can add a condition that applies to a relative (or absolute node) to any step of an XPath expression.
In this case:
//*[#id=substring-after(/Reference/#URI, '#')]
The //* matches all elements in the document. The part in [] is a condition. Inside the condition the part of the URI element of the root References node is taken, but ignoring the '#' (and anything before it).
Sample code, assuming you have loaded your XML into XPathDocument doc:
var nav = doc.CreateNavigator();
var found = nav.SelectSingleNode("//*[#id=substring-after(/Reference/#URI, '#')]");

If you have the value of the URI attribute in a variable you could use
myXmlDocument.DocumentElement.SelectSingleNode("//SOAP-ENV:Body[ID='pURI']")
where pURI is the value of the URI attribute and myXmlDocument is the Xml Document object

Something like this:
XmlDocument requestDocument = new XmlDocument();
requestDocument.LoadXml(yourXmlString);
String someXml = requestDocument.SelectSingleNode(#"/*[local-name()='Reference ']/*[local-name()='Body']").InnerXml;