Quick Tip: Using XPath to find nodes by attribute value

There are some things that I can just remember: phone numbers, locker combinations, and the like.  There are others that I have to lookup again and again and again.  XPath query syntax is one of the latter items.  Today, I’d like to talk a little about XPath queries.

The following XML contains a collection of “Friend” nodes.  Each node has attributes for name and birth date.  To keep this post to a reasonable length, I am going to limit the XML to 5 nodes.

  <Friend name=”Sam” birthMonth=”July” birthDay=”16″ birthYear=”1974″ />
  <Friend name=”Pam” birthMonth=”April” birthDay=”7″ birthYear=”1967″ />
  <Friend name=”George” birthMonth=”February” birthDay=”2″ birthYear=”1981″ />
  <Friend name=”John” birthMonth=”April” birthDay=”11″ birthYear=”1972″ />
  <Friend name=”Martha” birthMonth=”August” birthDay=”3″ birthYear=”1974″ />

After loading this XML into an XmlDocument, I can use XPath to query for the nodes of interest.

Since April is coming soon, I would like to get a list of my friends who has a birthday that month.  To do so, I query for Friend nodes where the value of the birthMonth attribute is “April”.

// load the xml file into an XmlDocument
XmlDocument doc = new XmlDocument();
XmlElement rootNode = doc.DocumentElement;

// select the friends who have a birthday in April
String query = “Friend[@birthMonth=’April’]”;
XmlNodeList friends = rootNode.SelectNodes(query);

foreach(XmlElement friend in friends)
    // TODO: display a reminder to buy a birthday card

When the above snippet is run, two friends are returned: Pam and John.  The application can retrieve the desired information from the node and display an appropriate reminder.

— DK

Edit: Fix grammar

This posting is provided “AS IS” with no warranties, and confers no rights.

Comments (0)

Skip to main content