Hi all,
I downloaded everything required to make apache client for a web
service. I am using "xerces" which is a JAXP compatible xml parser.
I developed a standalone application and it is working fine but when i
test an applet with same functionality something goes wrong.
When my applet is loaded it gives an exception given below
How does the program know about the xml parser i am using?
Where is it registered???
The Xmlparserutil is not able to find the xml parser it seems!!
Please......Help!!!!
om.ms.security.SecurityExceptionEx[soipapplet.start]: cannot access file
C:\WINNT\Java\lib\jaxp.properties
at com/ms/security/permissions/FileIOPermission.check
at com/ms/security/PolicyEngine.deepCheck
at com/ms/security/PolicyEngine.checkPermission
at com/ms/security/StandardSJcurityManager.chk
at com/ms/security/StandardSecurityManager.checkRead
tt java/io/File.exists
at javax/xml/parsers/FactoryFinder.find
at javax/xml/parsers/DocumentBuilderFactory.newInstance
/at org/apache/soap/util/xml/XMLParserUtils.refreshDocumentBuilderFactory
at org/apache/soap/util/xml/XMLParserUtils.<clinit>
at org/apache/soap/util/xml/XMLParserUtils.getXMLDocBuilder
at org/apache/soap/rpc/Call.<init>
at org/apache/soap/rpc/Call.<init>
at org/apache/soap/rpc/Call.<init>
at soapapplet.start
at com/ms/applet/AppletPanel.securedCall0
at com/ms/applet/AppletPanel.securedCall
at com/ms/applet/AppletPanel.processSentEvent
at com/ms/applet/AppletPanel.run
at java/lang/Thread.run
