Note that without the generated code or jar file you will not be able to access the values. You can parse the tree though.
________________________________ From: Andrew Mansfield [mailto:[EMAIL PROTECTED] Sent: 15 September 2008 14:59 To: user@xmlbeans.apache.org Subject: RE: Very simple question (I think) I just ran the following Junit test with no problems... public void test() throws XmlException, IOException{ String xmltext = new String("<test><element1>testvalue</element1></test>"); XmlObject xmlObject = XmlObject.Factory.parse(new ByteArrayInputStream(xmltext.getBytes())); Node rootnode = xmlObject.getDomNode().getFirstChild(); assertEquals("test",rootnode.getNodeName()); } ________________________________ From: Nicolai Odum [mailto:[EMAIL PROTECTED] Sent: 15 September 2008 14:41 To: user@xmlbeans.apache.org Subject: RE: Very simple question (I think) Sorry bad example It's was just suppose to be psudo code String xml = "big xml document"; I have tried it with many big valid xml documents - without luck. /Nicolai "Andrew Mansfield" <[EMAIL PROTECTED]> 15-09-2008 15:36 Please respond to user@xmlbeans.apache.org To <user@xmlbeans.apache.org> cc Subject RE: Very simple question (I think) I think you need to parse a valid source document first. Then you can get access to the underlying XmlObject. Regards, ________________________________ From: Nicolai Odum [mailto:[EMAIL PROTECTED] Sent: 15 September 2008 14:32 To: user@xmlbeans.apache.org Subject: Very simple question (I think) Hello On xmlbeans.apache.org <http://xmlbeans.apache.org/docs/2.0.0/guide/conGettingStartedwithXMLBea ns.html> it says that XMLBeans provide <quote> It provides a familiar Java object-based view of XML data without losing access to the original, native XML structure </quote> I am using XMLBeans on a xml structure that is signed with a hash value so *nothing* must change before the xml is invalid. The question is: *HOW* do i get access to the original, native XML structure. because this doesn't work String xml = "big xml document"; XmlObject parse = XmlObject.Factory.parse(xml); String newXml = parse.xmlText(); if (xml.equals(newXml) System.out.println("jubii"); else System.out.println("damn"); Please help :-) Cheers Nicolai ________________________________ This email with all information contained herein or attached hereto may contain confidential and/or privileged information intended for the addressee(s) only. If you have received this email in error, please contact the sender and immediately delete this email in its entirety and any attachments thereto. (W1) ________________________________ This email with all information contained herein or attached hereto may contain confidential and/or privileged information intended for the addressee(s) only. If you have received this email in error, please contact the sender and immediately delete this email in its entirety and any attachments thereto. (W1) -------------------------------------------------------------------------- This email with all information contained herein or attached hereto may contain confidential and/or privileged information intended for the addressee(s) only. If you have received this email in error, please contact the sender and immediately delete this email in its entirety and any attachments thereto. (W1)
