CODEWITHSUNDEEP
c#xmlcollectionsxsdxmlserializer
Jarlaxle
Jarlaxle

Reputation: 901

XML with collections and XmlSerializer

I have an XSD:

<xs:complexType name="rootType">    
  <xs:sequence>
    <xs:element name="foo" type="fooType" minOccurs="1" maxOccurs="unbounded"/>
    <xs:element name="bar" type="barType" minOccurs="1" maxOccurs="unbounded"/>
  </xs:sequence>    
</xs:complexType>
<!-- SKIPPED -->
<xs:element name="root" type="rootType"></xs:element>

I have an XML built using this XSD:

<?xml version="1.0" encoding="utf-8"?>
<root>
  <foo><!-- SKIPPED --></foo>
  <foo><!-- SKIPPED --></foo>
  <foo><!-- SKIPPED --></foo>
  <bar><!-- SKIPPED --></bar>
  <bar><!-- SKIPPED --></bar>
  <bar><!-- SKIPPED --></bar>
  <bar><!-- SKIPPED --></bar>
</root>

Now I want to serialize/deserialize this XML using XmlSerializer. I have C# classes:

public class fooType
{
    public string element { get; set; }
}

public class barType
{
    public string element { get; set; }
}

public class rootType
{
    public fooType[] foos { get; set; }
    public barType[] bars { get; set; }
}

There were some XML-related attributes, like XmlElementAttribute, but I omit them in example above for simplicity.

Now please take a look at rootType class definition. Here we have two properties foos and bars. They will be serialized like root elements for fooType[] and barType[] arrays:

<?xml version="1.0" encoding="utf-8"?>
<root>
  <foos>
    <foo><!-- SKIPPED --></foo>
    <foo><!-- SKIPPED --></foo>
    <foo><!-- SKIPPED --></foo>
  </foos>
  <bars>
    <bar><!-- SKIPPED --></bar>
    <bar><!-- SKIPPED --></bar>
    <bar><!-- SKIPPED --></bar>
    <bar><!-- SKIPPED --></bar>
  </bars>
</root>

But this is not what I want. How to serialize them according to XSD and example at the beginning of this post?

Upvotes: 0

Views: 565

Answers (2)

Christian Hayter
Christian Hayter

Reputation: 31071

I find that the quickest way of working out how to design your classes to match a specific schema is to run the xsd.exe tool backwards. Tell it to generate classes from your schema, and compare them with your hand-written classes to see where you went wrong. It's a great way of gaining experience in how the serializer works.

Upvotes: 1

Erik Philips
Erik Philips

Reputation: 54638

Based on your criteria, I would Implement IXmlSerializable on rootType. There is a great read here on SO on how to create the XML you are looking for.

Proper way to implement IXmlSerializable?

Upvotes: 1

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