Adding element to XML using linq to XML

Question

i have this piece of code which i use to add some elements:

  string xmlTarget = string.Format(@"<target name='{0}' type='{1}' layout='${{2}}'  />",
                                                new object[] { target.Name, target.Type, target.Layout });
            Configuration configuration = ConfigurationManager.OpenExeConfiguration(ConfigurationUserLevel.None);
            var xmlDoc = XElement.Load(configuration.FilePath);
            var nlog = xmlDoc.Elements("nlog");

            if (nlog.Count() == 0)
            {
                return false;
            }
            xmlDoc.Elements("nlog").First().Elements("targets").First().Add(xmlTarget);
            xmlDoc.Save(configuration.FilePath,SaveOptions.DisableFormatting);
            configuration.Save(ConfigurationSaveMode.Modified);
            ConfigurationManager.RefreshSection("nlog");
            return true;

it supposed to add a target to the xml , problem is it replace "<" with "<" and ">" with ">" which mess up my xml file.

how do i fix this ?

Note please dont pay attention to nlog, i`m concerned about the linqtoxml problem.

Quick note: there's an easy way to get us to not pay attention to particular parts of the code... trim them out of the post. (As an aside, using if (!nlog.Any()) is better than using if (nlog.Count() == 0).) — Jon Skeet
– Jon Skeet, Commented Sep 26, 2011 at 11:53

Jon Skeet · Accepted Answer · 2011-09-26 11:51:47Z

4

You're currently adding a string. That will be added as content. If you want to add an element, you should parse it as such first:

XElement element = XElement.Parse(xmlTarget);

Or preferrably, construct it instead:

XElement element = new XElement("target",
    new XAttribute("type", target.Name),
    new XAttribute("type", target.Type),
    // It's not clear what your format string was trying to achieve here
    new XAttribute("layout", target.Layout));

Basically, if you find yourself using string manipulation to create XML and then parse it, you're doing it wrong. Use the API itself to construct XML-based objects.

answered Sep 26, 2011 at 11:51

Jon Skeet

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