Can I fail to deserialize with XmlSerializer in C# if an element is not found?

问题

I am using XmlSerializer to write and read an object to xml in C#. I currently use the attributes XmlElement and XmlIgnore to manipulate the serialization of the object.

If my xml file is missing an xml element that I require, my object still deserializes (xml -> object) just fine. How do I indicate (preferably via Attributes) that a certain field is \"required\"?

Here is a sample method of what I am using currently:

[XmlElement(ElementName=\"numberOfWidgets\")]
public int NumberThatIsRequired {
    set ...;
    get ...;
}

My ideal solution would be to add something like an XmlRequired attribute.

Also, is there a good reference for what Attributes are available to manipulate the behavior of XmlSerializer?

回答1:

I've got an answer for the second part: "Attributes that control XML serialization".

Still investigating the first part...

EDIT: I strongly suspect you can't do this through XML deserialization itself. I've just run xsd.exe on a sample schema which includes a required attribute - and it's exactly the same if the attribute is marked as being optional. If there were a way of requiring properties to be set, I'd expect it to be implemented in that case.

I suspect you've basically got to just validate your tree of objects after deserializing it. Sorry about that...

回答2:

The only way I've found to do this is via XSD. What you can do is validate while you deserialize:

static T Deserialize<T>(string xml, XmlSchemaSet schemas)
{
    //List<XmlSchemaException> exceptions = new List<XmlSchemaException>();
    ValidationEventHandler validationHandler = (s, e) =>
    {
        //you could alternatively catch all the exceptions
        //exceptions.Add(e.Exception);
        throw e.Exception;
    };

    XmlReaderSettings settings = new XmlReaderSettings();
    settings.Schemas.Add(schemas);
    settings.ValidationType = ValidationType.Schema;
    settings.ValidationEventHandler += validationHandler;

    XmlSerializer serializer = new XmlSerializer(typeof(T));
    using (StringReader sr = new StringReader(xml))
        using (XmlReader books = XmlReader.Create(sr, settings))
           return (T)serializer.Deserialize(books);
}

回答3:

For extensibility reasons, XmlSerializer is very forgiving when it comes to deserialization; things like [DefaultValue], ShouldSerialize{Foo} and {Foo}Specified are mainly used during serialization (the exception being {Foo}Specified, which is set during deserialization as well as queried during serialization).

As such; there isn't an easy way to do this, unless you implement IXmlSerializable and do it yourself. damagednoob shows an xsd option, which is also an option.

来源：https://stackoverflow.com/questions/259726/can-i-fail-to-deserialize-with-xmlserializer-in-c-sharp-if-an-element-is-not-fou

标签

xml

xml-serialization

.net-attributes