XmlSerializer List Item Element Name

Question

I have a class PersonList

[XmlRoot(\"Persons\")]
PersonList : List

when I serialize this to XML, by default it wi

Answer 1

There is another alternative. You can always implement IXmlSerializable in collections (and any other class or struct) to fully control how items are written or read. Many of you will know that already, it's not generally preferred of course as you may end-up writing out "boiler-plate" code by hand which should really be automatic logic specified with attributes.

For collections which must match a sensible schema it's justifiable. Because the framework has a hard limitation here and the existing item type's serialization code does not have to be duplicated when done properly; i.e. Do not re-write the item serialization in the collection code, just create/call a child XmlSerializer inside your ReadXml/WriteXml implementation.

Once consequence of using IXmlSerializable is it does not allow you to apply the XmlTypeAttribute (throws a runtime error telling you only XmlRootAttribute may be used). So instead apply the XmlSchemaProviderAttribute and return the same qualified name you would have put in the XmlTypeAttribute. The old GetSchema method should return null anyway as it was only a reserved method (according to MSDN), probably because they forgot to include the ability to specify a different namespace. Personally I use the same "GetSchema" method name in my XmlSchemaProviderAttribute so it appears as a complete override next to the legacy placeholder GetSchema method.

Of course, the best solution would be if Microsoft would allow us to apply the XmlArrayItemAttribute to collection/list classes and use that in the XmlSerializer. By default it uses the XML type element name in collections, which I feel is a bug because it should be the XML root name when specified or class name when not.

When I get time I'll come back and add an example. For now take a look at the MSDN documentation examples of IXmlSerializable and XmlSchemaProviderAttribute.

http://msdn.microsoft.com/en-us/library/system.xml.serialization.ixmlserializable(v=vs.110).aspx

http://msdn.microsoft.com/en-us/library/system.xml.serialization.xmlschemaproviderattribute(v=vs.110).aspx

Answer 2

If you don't have access to the source for the Human class (in which case, setting XmlRoot is not possible), you can create an XmlElementAttribute, then add it to an XmlAttributeOverride and use that when creating an instance of your XmlSerializer. See this MSDN article for more details.