CODEWITHSUNDEEP
xsdxml-validation
akarnokd
akarnokd

Reputation: 69997

Declaring an element with a type xml schema

How can I set an xsd:element type to be the XMLSchema (which defines the structure of XSD files)? Is it even possible?

For example, I need a XML file which lists multiple XSDs under its root element:

<schemas xmlns:xs=''>
  <xs:schema...>
    <xs:element name='...'/>
  </xs:schema>
  <xs:schema...>
  </xs:schema>
</schemas>

The schema of this XML would look like this:

<xs:schema xmlns:xs=''>
  <xs:element name='schemas'>
    <xs:complexType>
      <xs:sequence>
        <xs:element name='schema' type='xs:schema' 
            minoccurs='0' maxoccurs='unbounded'/>
      </xs:sequence>
    </xs:complexType>
  </xs:element>
</xs:schema>

Of course, there is no such type as xs:schema. How can I make this work?

Upvotes: 0

Views: 438

Answers (1)

ColdFusion
ColdFusion

Reputation: 2531

Yes, it is quite possible. Here's how:

<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema">

  <!-- 
    Any XML schema processor does know XML schema language,
    but it is not supposed to know an XML schema for it.
    You need to import it!
  -->
  <xs:import namespace="http://www.w3.org/2001/XMLSchema"
             schemaLocation="http://www.w3.org/2001/XMLSchema.xsd"/>

  <xs:element name="schemas">
    <xs:complexType>
      <xs:sequence>
        <!-- 
          You don't need an 'xs:schema' type. Rather you just need
          to reference an already existing 'xs:schema' element
        -->
        <xs:element minoccurs="0" maxoccurs="unbounded" ref="xs:schema"/>
      </xs:sequence>
    </xs:complexType>
  </xs:element>

</xs:schema>

You may look also in the XML schema for XSLT: http://www.w3.org/2007/schema-for-xslt20.xsd. They do the same in the definition of <xsl:import-schema> element.

Upvotes: 3

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