I have the following (as an example) XML file and XSD.
<?xml version="1.0" encoding="utf-8" ?>
<foo>
<DateVal>2010-02-18T01:02:03</DateVal>
<TimeVal>PT10H5M3S</TimeVal>
</foo>
and
version="1.0" encoding="utf-8"?>
<xs:schema attributeFormDefault="unqualified" elementFormDefault="qualified" xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xs:element name="foo">
<xs:complexType>
<xs:sequence>
<xs:element name="DateVal" type="xs:dateTime" />
<xs:element name="TimeVal" type="xs:duration" />
</xs:sequence>
</xs:complexType>
</xs:element>
</xs:schema>
Then the following C# code:
static void Main(string[] args)
{
XmlDocument xd = new XmlDocument();
XmlSchema xs;
using (var fs = File.OpenRead(FilePath + "SimpleFields.xsd"))
{
xs = XmlSchema.Read(fs, null);
}
xd.Schemas.Add(xs);
xd.Load((FilePath + "SimpleFields.xml"));
xd.Validate(null);
var el_root = xd.DocumentElement;
var el_date = (XmlElement)el_root.SelectSingleNode("./DateVal");
//WANTED: el_date.Value = 2010-02-18 01:02:03 (as a DateTime Object)
//ACTUAL: el_date.InnerText="2010-02-18T01:02:03"
var el_duration = (XmlElement)el_root.SelectSingleNode("./TimeVal");
//WANTED: el_date.Value = 10 hours, 5 minutes, 3 seconds (as a TimeSpan Object)
//ACTUAL: el_date.InnerText="PT10H5M3S"
Console.WriteLine("DONE");
Console.ReadLine();
}
How can I read the data as strongly typed objects ?
I will be targetting a WindowsMobile device, but this shouldn't need to affect the answer too much. (can be .NET 2.0 or 3.5 ... Not sure if Sstem.Xml.Linq will help or not)
You need to do two steps:
1) Take your XML schema file and run it through the xsd.exe
utility (which comes with the Windows SDK - it's in C:\Program Files\Microsoft SDKs\Windows\v6.0A\Bin\
or some similar path. This can turn the XSD file into a C# class:
xsd /c yourfile.xsd
This should give you a file yourfile.cs
which contains a class representing that XML schema.
2) Now, armed with that C# class, you should be able to just deserializing the XML file into an instance of your new object:
XmlSerializer ser = new XmlSerializer(typeof(foo));
string filename = Path.Combine(FilePath, "SimpleFields.xml");
foo myFoo = ser.Deserialize(new FileStream(filename, FileMode.Open)) as foo;
if (myFoo != null)
{
// do whatever you want with your "foo"
}
That's about as simple as it gets! :-)