|
Hi,
XML Schema does not define a root element. You can
use any of the top level elements as root element for your
document.
For instance both:
<?xml version="1.0"
encoding="UTF-8"?>
<comment xmlns="http://www.example.com/IPO"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.example.com/IPO
ipo.xsd">
...
</comment>
and
<?xml version="1.0"
encoding="UTF-8"?>
<purchaseOrder xmlns="http://www.example.com/IPO"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.example.com/IPO
ipo.xsd">
....
</purchaseOrder>
are valid documents against ipo.xsd
schema.
Regards,
George
-------------------------------------------------------------
George
Cristian Bina
<oXygen/> XML Editor - http://www.oxygenxml.com/
----- Original Message -----
Sent: Friday, October 10, 2003 1:24
PM
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
| 
Cart
