Generating XSD for serialized XML

Question

Currently, I have an xml file that looks like this...


<ArrayOfService>
    <Service>
        <Name>
            Something
        </Name>
        <Id>
            8003
        </Id>
    </Service>
</ArrayOfService>

This is automatically generated from a class that looks like this...


public class Service{
    public string Name;
    public int Id;

    public Service(){
    }
}

To turn the class into XML, I use...


XmlSerializer xs = new XmlSerializer( typeof(Service) );
xs.Serialize( context.Response.OutputStream, FunctionReturnsTypeService() );

Is there any way to also automatically generate an XSD like this?

EDIT:

Also, is there any way to add this schema to the xml as I'm serializing it?

what do you mean "add this schema to the XML"? What do you want the output to look like, to contain? — Cheeso, Jun 09 '09 at 15:08
I want the generated xml to have something like this at the top: — apandit, Jun 09 '09 at 15:10
xmlns="foo" indicates a namespace, not a schema. If you want a specific xml namespace to be used when serializing a type, then you can use [XmlType(Namespace="Foo")] as an attribute on the type. — Cheeso, Jun 09 '09 at 15:15
I think what you may want is xsi:schemaLocation, in which case, see here: http://www.tech-archive.net/Archive/DotNet/microsoft.public.dotnet.xml/2006-12/msg00040.html — Cheeso, Jun 09 '09 at 15:19
Add that link to your answer, I'll accept it. Looks like what i needed. — apandit, Jun 09 '09 at 15:25

Cheeso · Accepted Answer · 2009-06-09T15:32:15.147

2

The xsd.exe tool (%netsdk20%\bin\xsd.exe) infers a type from an XML document.

(You can also use the /c option to generate classes from an xml doc or schema.)

If you want to embed a reference to a schema into an XML doc, then see here: http://www.tech-archive.net/Archive/DotNet/microsoft.public.dotnet.xml/2006-12/msg00040.html

Summary:
decorate a member of your type with the XmlAttribute attribute, specifying "schemaLocation" as the name of the attr, and "http://www.w3.org/2001/XMLSchema-instance" as the namespace for that attribute. As this example in C#

[System.Xml.Serialization.XmlAttributeAttribute("schemaLocation",
    Namespace = System.Xml.Schema.XmlSchema.InstanceNamespace)]
private string xsiSchemaLocation = "YourSchema.xsd";

edited Jun 09 '09 at 15:32

answered Jun 09 '09 at 14:59

Cheeso

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the XSD tool works, but not the /c option (that's for generating classes). I just typed xsd myfile.xml and it generated the schema. Awesome. Now, for the second part of my question.. :D – apandit Jun 09 '09 at 15:06
For on the fly schema generation, http://stackoverflow.com/questions/336988/xml-serialization-and-schema-without-xsd-exe – apandit Jun 09 '09 at 15:26
The field xsiSchemaLocation needs to be public. – row1 Jul 29 '11 at 03:00

Generating XSD for serialized XML

1 Answers1