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str.Append("</b><br />"); break; } ... Note that the values for the NodeType property are almost the same, except for the enumeration name, which is XPathNodeType instead of XmlNodeType. That s because the XPathNavigator uses a smaller, more streamlined set of nodes. One of the nodes it doesn t support is the XmlDeclaration node type. The function checks if the current node has any attributes. If so, it moves to the first one with a call to MoveToFirstAttribute() and loops through all the attributes until the MoveToNextAttribute() method returns false. At that point it returns to the parent node, which is the node originally referenced by the object. Here s the code that carries this out: ... if (xnav.HasAttributes) { xnav.MoveToFirstAttribute(); do { str.Append(indent); str.Append(" - Attribute: <b>"); str.Append(xnav.Name); str.Append("</b> Value: <b>"); str.Append(xnav.Value); str.Append("</b><br />"); } while (xnav.MoveToNextAttribute()); // Return to the parent. xnav.MoveToParent(); } ... The function does a similar thing with the child nodes by moving to the first one with MoveToFirstChild() and recursively calling itself until MoveToNext() returns false, at which point it moves back to the original node, as follows: ... if (xnav.HasChildren) { xnav.MoveToFirstChild(); do { str.Append(GetXNavDescr(xnav, level+1)); } while (xnav.MoveToNext()); // Return to the parent. xnav.MoveToParent(); } return str.ToString(); } This code produces almost the same output as shown in Figure 12-2.

This basic decorator will allow a view to have a template loaded, rendered and returned automatically. Simply provide a template name to the decorator and return a dictionary, and the decorator does the rest. Moving on to the other side of the problem, the decorator needs to be able to return a string that can be read by JavaScript, containing the same content that would otherwise be provided by the template. Django comes built-in with a copy of ,2 a library designed for converting Python objects into JavaScript Object Notation (JSON),3 so it can be easily consumed by code inside the Web browser. It even has a custom encoder for use with , designed to encode some of the particular types Django uses. This encoder can be used to transmit the dictionary s content directly to the browser, bypassing the template entirely. Combining it with covered in 7 allows the decorator to decide which path to take, completing the equation.

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In some situations, you don t need to process the entire XML document. Instead, you need to extract a single piece of information. If you know the element name, you can use the XmlDocument.GetElementsByTagName() method, which searches an entire document and returns an XmlNodeList that contains all the matching XmlNode objects.

For example, the following code retrieves the title of each DVD in the document: // Load the XML file. string xmlFile = Server.MapPath("DvdList.xml"); XmlDocument doc = new XmlDocument(); doc.Load(xmlFile); // Find all the <Title> elements anywhere in the document. StringBuilder str = new StringBuilder(); XmlNodeList nodes = doc.GetElementsByTagName("Title"); foreach (XmlNode node in nodes) { str.Append("Found: <b>"); // Show the text contained in this <Title> element. str.Append(node.ChildNodes[0].Value); str.Append("</b><br />"); } XmlText.Text = str.ToString(); Figure 12-3 shows the result of running this code in a web page.

Figure 12-3. Searching for information in an XML document You can also search portions of an XML document by using the method XmlElement.GetElementsByTagName() on a specific element. In this case, the XmlDocument searches all the descendant nodes looking for a match. To use this method, first retrieve an XmlNode that corresponds to an element and then cast this object to an XmlElement. The following example demonstrates how to use this technique to find the stars of a specific movie: // Load the XML file. string xmlFile = Server.MapPath("DvdList.xml"); XmlDocument doc = new XmlDocument(); doc.Load(xmlFile); // Find all the <Title> elements anywhere in the document. StringBuilder str = new StringBuilder(); XmlNodeList nodes = doc.GetElementsByTagName("Title"); foreach (XmlNode node in nodes) { str.Append("Found: <b>");

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// Show the text contained in this <Title> element. string name = node.ChildNodes[0].Value; str.Append(name); str.Append("</b><br />"); if (name == "Forrest Gump") { // Find the stars for just this movie. // First you need to get the parent node // (which is the <DVD> element for the movie). XmlNode parent = node.ParentNode; // Then you need to search down the tree. XmlNodeList childNodes = ((XmlElement)parent).GetElementsByTagName("Star"); foreach (XmlNode childNode in childNodes) { str.Append("   Found Star: "); str.Append(childNode.ChildNodes[0].Value); str.Append("<br />"); } } } XmlText.Text = str.ToString(); Figure 12-4 shows the result of this test.

Figure 12-4. Searching portions of an XML document The code you ve seen so far assumes that none of the elements has a namespace. More sophisticated XML documents will always include a namespace and may even have several of them. In this situation, you can use the overload of the method XmlDocument.GetElementsByTagName(), which requires a namespace name as a string argument, as shown here: // Retrieve all <order> elements in the OrderML namespace. XmlNodeList nodes = doc.GetElementsByTagName("order", "http://mycompany/OrderML"); Additionally, you can supply an asterisk (*) for the element name if you want to match all tags in the specified namespace:

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