如何在xml中获取特定类型的节点属性？

Question

I am using the following code:

System.Xml.XmlDocument document = new System.Xml.XmlDocument();
document.Load(@"D:\Files\OCR\" + FileUpload1.FileName + ".xml");

if (document.HasChildNodes)
{
    StringBuilder sb = new StringBuilder();
    StringBuilder positions = new StringBuilder();
    XmlElement root = document.DocumentElement;
    XmlNodeList nodes = document.DocumentElement.SelectNodes("//char[@confidence]");
}

The problem is that document.DocumentElement.SelectNodes("//char[@confidence]") returns null.

When I write the following code the result is shown.

int nodesCount = Document.DocumentElement.ChildNodes[0].ChildNodes.Count;

How do I count all nodes with attributes confidence?

Answer 1

You could use XDocument and some effective LINQ:

XDocument doc = XDocument.Load(@"D:\Temp\file.xml");
int count = doc.Root.Descendants().Count(e => e.Attribute("confidence") != null);
Console.Write("Count:" + count);
Console.Read();

Output: 4

And my file.xml contains the following:

<something>
    <char confidence="1">
    </char>
    <char confidence="2">
    </char>
    <char confidence="3">
    </char>
    <notchar confidence="1">
    </notchar>
</something>

The above code checks all descendants for the attribute "confidence". If you want only elements that have the name "char", you can use the following:

int count = doc.Root.Descendants().Count(e => e.Name == "char" && e.Attribute("confidence") != null);

Output: 3