Question

What is the fastest method to retrieve an XML node? I have an application that need the functionality of replace a specific node, when the document is small is fast but soon get bigger then it takes some several second to do the replacement. This is the method, I just do a brute force comparison that sucks real bad in that scenario.

public bool ReplaceWithAppendFile(string IDReplace)
{
    XElement UnionElement = (from sons in m_ExtractionXmlFile.Root.DescendantsAndSelf()
                             where sons.Attribute("ID").Value == IDReplace
                             select sons).Single();
    UnionElement.ReplaceWith(m_AppendXmlFile.Root.Elements());
    m_ExtractionXmlFile.Root.Attribute("MaxID").Value =
        AppendRoot.Attribute("MaxID").Value;
    if (Validate(m_ExtractionXmlFile, ErrorInfo))
    {
        m_ExtractionXmlFile.Save(SharedViewModel.ExtractionFile);
        return true;
    }
    else
    {
        m_ExtractionXmlFile = XDocument.Load(SharedViewModel.ExtractionFile);
        return false;
    }
}

Answer 1

Try using XPath:

string xPath = string.Format("//*[@id='{0}']", IDReplace);
XElement UnionElement = m_ExtractionXmlFile.XPathSelectElement(xPath);

You may refer to Finding Elements by Attributes in a DOM Document Using XPath for more examples.

P.S. It is considered good convention to start names of parameters and local variables in lowercase. Thus, use idReplace and unionElement rather than the above.