REST web service set up

Question

Hello,
Need help setting up an xpath for a rest web service
response path:  //m:properties
REST output parameters
name:  project   path: /d:WO_PREFIX
When I try to create a Form and Report I get an error using the defaults:
ORA-20001: Unable to create form on table. ORA-00907: missing right parenthesis
What would be the correct setup for this to work????
Here is an example of my response
<?xml version="1.0" encoding="utf-8"?>
<feed xml:base="http://xxx.xx.com/EF/odata" xmlns="http://www.w3.org/2005/Atom" xmlns:d="http://schemas.microsoft.com/ado/2007/08/dataservices" xmlns:m="http://schemas.microsoft.com/ado/2007/08/dataservices/metadata" xmlns:georss="http://www.georss.org/georss" xmlns:gml="http://www.opengis.net/gml">
  <id>http://schemas.datacontract.org/2004/07/</id>
  <title />
  <updated>2013-08-28T17:51:58Z</updated>
  <link rel="self" href="http://xxx.xx.com/EF/odata/PROJECTS" />
  <entry>
    <id>http://xxx.xxx.com/EF/odata/PROJECTS('ZZZ')</id>
    <category term="WOSApi.Models.PROJECT" scheme="http://schemas.microsoft.com/ado/2007/08/dataservices/scheme" />
    <link rel="edit" href="http://xxx.xx.com/EF/odata/PROJECTS('ZZZ')" />
    <link rel="self" href="http://xxx.xx.com/EF/odata/PROJECTS('ZZZ')" />
    <title />
    <updated>2013-08-28T17:51:58Z</updated>
    <author>
      <name />
    </author>
    <content type="application/xml">
      <m:properties>
        <d:WO_PREFIX>ZZZ</d:WO_PREFIX>
      </m:properties>
    </content>
  </entry>
</feed>

Answer 1

Use Xmlspy or some editor to modify or edit the WSDL as per need.
elementFormDefault="qualified"
This means that all the elements in the XML document must have namespace variable prefixed with them.
elementFromDefault="unqualified"
This means that not all the elements in the XML document requires namespace variable prefixed with them. Only the root element has to be prefixed with namespace variable.