如何获得xml节点的同级

Question

I have an xml file as below

<Games>
   <Game>
      <name>Tzoker</name>
      <file>tzoker1</file>
   </Game>
   <Game>
      <file>lotto770</file>
   </Game>
   <Game>
      <name>Proto</name>
      <file>proto220</file>
   </Game>
</Games>

I want to get the values of name and file items for every Game node. It is easy by using this query.

string query = String.Format("//Games/Game");
 XmlNodeList elements1 = xml.SelectNodes(query);
 foreach (XmlNode xn in elements1)
 {
   s1 = xn["name"].InnerText;
   s2 = xn["file"].InnerText;
 }

The problem is that there are some nodes that they don't have the name item. So the code above doesn't work. I have solved the problem by using the following code

string query = String.Format("//Games/Game/name");
XmlNodeList elements1 = xml.SelectNodes(query);
foreach (XmlNode xn in elements1)
{
  s1 = xn.InnerText;
  string query1 = String.Format("//Games/Game[name='{0}']/file", s1);
  XmlNodeList elements2 = xml.SelectNodes(query1);
  foreach (XmlNode xn2 in elements2)
  {
    s2 = xn2.InnerText;
  }
}

The problem is that there is a case that two or more nodes have the same name value. So, the s2 variable will get the file value of the last node that the loop finds. So, I would like to find a way to get the sibling file value of the current name item. How could I do it? I try do move to the parent node of the current node and then to move to the file item but without success by using the following code.

string query = String.Format("//Games/Game/name");
XmlNodeList elements1 = xml.SelectNodes(query);
foreach (XmlNode xn in elements1)
{
   s1 = xn.InnerText;
   string query1 = String.Format("../file");
   XmlNodeList elements2 = xml.SelectNodes(query1);
   foreach (XmlNode xn2 in elements2)
   {
     s2 = xn2.InnerText;
   }
}

I hope there is a solution.

Answer 1

If I understand you correctly you want to find all games that have a name. You can do that using XPath. Here is a solution that uses LINQ to XML. I find that easier to work with than XmlDocument :

var xDocument = XDocument.Parse(xml);
var games = xDocument
  .Root
  .XPathSelectElements("Game[child::name]")
  .Select(
    gameElement => new {
      Name = gameElement.Element("name").Value,
      File = gameElement.Element("file").Value
    }
  );

The XPath to select all <Game> elements that have a <name> child element is Game[child::name] .

Answer 2

You can use Game[name] to filter Game elements to those with child element name . This is possible because child:: is the default axes which will be implied when no explicit axes mentioned. Extending this further to check for child element file as well, would be as simple as Game[name and file] :

string query = String.Format("//Games/Game[name]");
XmlNodeList elements1 = xml.SelectNodes(query);
foreach (XmlNode xn in elements1)
{
   s1 = xn["name"].InnerText;
   s2 = xn["file"].InnerText;
}

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Now to answer your question literally, you can use following-sibling:: axes to get sibling element that follows current context element. So, given the context element is name , you can do following-sibling::file to return the sibling file element.

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Your attempt which uses ../file should also work. The only problem was, that your code executes that XPath on xml , the XmlDocument , instead of executing it on current name element :

XmlNodeList elements2 = xn.SelectNodes("../file");