As they said, W3C XML Schemas can be used to
validate many different XML documents. Any "top-level" or "global"
<xsd:element> tag can be the document element of a valid XML
document.
As for your project, take a look at how we're
handling that with xmlArchitect (http://www.sysonyx.com/products/xmlarchitect).
If the schema only has one element, our instance tree just shows that element.
If it has more than one, a drop-down appears above the tree, allowing the user
to select which element they consider the document element.
Otherwise, to avoid this in general schema-design,
only have one global element, make the rest complexType or simpleType nodes. One
should also note that for a schema to be valid, you have to have a least one
global element (as opposed to a bunch of type nodes).
Bryce K. Nielsen
SysOnyx, Inc. (http://www.sysonyx.com)
Makers of xmlLinguist, the Text-to-XML
Translator
(http://www.sysonyx.com/products/xmllinguist)
----- Original Message -----
Sent: Friday, October 10, 2003 5:24
AM
Subject: Help for finding root
node from XML Schema
Hello all,
I am new to XML and XML Schema. I am working on a
project which involves visual display of XML Schema. For parsing XML
Schema I am using MSXML 4.0
My problem is How to get element which will
become root element of XML document which uses this schema.
e.g.
PO.xsd
<xsd:schema
xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns=""
elementFormDefault="qualified"
finalDefault="restriction">
<xsd:element name="purchaseOrder"
type="PurchaseOrderType" />
<xsd:element name="comment"
type="xsd:string"/>
<xsd:complexType
name="PurchaseOrderType"> <xsd:sequence> ... </xsd:sequence> <xsd:attribute
name="orderDate"
type="xsd:date"/> </xsd:complexType>
</xsd:schema>
When I parse above schema first call to
get_childs returns two childs 1 ) comment 2) purchaseOrder.
So how to determine which is root?
Thanks & Regards, Priti
Patil
|