Problem in accessing XML files using Web Start

Hello,
I am trying to run a swing based application, for which I am using Java Web Start. I have 3 jar files. One is ma_ui.jar which contains all my class files. 2nd is config.jar which contains all my xml and other image files. 3rd is classes111.zip for database access. When I try to run this application, the error it shows is application.xml is not in the classpath. But I am sure that application.xml is there in the config.jar. Can anybody help me out.
Thanks and Regards
Suneet

It would help if you include a copy of your jnlp with this post.

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I am reposting it as I was told to format the code and send it again.
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Xml
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`
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Sax.endValidation();
Sax.endDocument();
* I HAVE CHANGED MY AS PER YOUR SUGGESTION, NOW SAME RESULT HAS COMING.
* I AM NOT ABLE TO GET THE EXACT OUTPUT.,WHY THAT ?
Thanks,
JavaImran{code}

How can we get tag of XML file using SAX

Hi ,
I'm parsing one SAX parser , I'have almost done this parsing. i have faced problem for one case, i'e how can we get tag from XML file using SAX parser?
XML file is
<DFProperties>
<AccessType>
<Get/>
</AccessType> <Description>
gdhhd
</Description>
<DFFormat>
<chr/>
</DFFormat>
<Scope>
<Permanent/>
</Scope>
<DFTitle>gsgd</DFTitle>
<DFType>
<MIME>text/plain</MIME>
</DFType>
</DFProperties>
I want out like GET and Permanent... means this one tag which is present inside of another tag.
Handler class like
public void startElement(String namespaceURI, String localName,
               String qName, Attributes atts) throws SAXException {
if(_ACCESSTYPE.equals(localName)){
               accessTypeElement=ACCESSTYPE;
public void characters(char[] ch, int start, int length)
               throws SAXException {
if (_ACCESSTYPE.equals(_accessTypeElement)) {
               String strValue = new String(ch, start, length);
               System.out.println("Accestype-----------------------------> " + strValue);
               //System.out.println(" " + strValue);
public void endElement(String namespaceURI, String localName, String qName)
               throws SAXException {
if (_ACCESSTYPE.equals(localName)) {
               _accessTypeElement = "";
. please any body help me

Hi ,
I have one problem,Please help me.
1. How can I'll identify where exactly my Node is ended,means how how can we find corresponding nodename? in partcular place
<Node> .............starttag1
<NodeName>Test</NodeName>
<Node>................starttag2
<nodeName>test1</NodeName>
</Node>..................endtag2
<Node>.....................starttag3
<NodeName><NodeName>
<Node> .........................starttag4
<NodeName>test4</NodeName>
</Node>.......enddtag4
</Node>...........end tag3
</Node>............endtag1
my code is below
private final String _NODENAME = "NodeName";
private final String _NODE = "Node";
private String _nodeElement = "";
     private String _NodeNameElement = "";
public void startElement(String namespaceURI, String localName,
               String qName, Attributes atts) throws SAXException {
if (_NODENAME.equals(localName)) {
               NodeNameElement = NODENAME;
if(_NODE.equals(localName)){
     System.out.println("start");
     if (_NODENAME.equals(localName)) {
               NodeNameElement = NODENAME;
public void characters(char[] ch, int start, int length)
               throws SAXException {
if (_NODENAME.equals(_NodeNameElement)) {
               String strValue = new String(ch, start, length);
               String sttt=strValue;
               System.out.println("NODENAME: ************* " + strValue);
if(_NODE.equals(_nodeElement)){
               if (_NODENAME.equals(_NodeNameElement)) {
                    String strValue = new String(ch, start, length);
                    String sttt=strValue;
                    System.out.println("nodevalue********** " + strValue);
public void endElement(String namespaceURI, String localName, String qName)
               throws SAXException {
if (_NODENAME.equals(localName)) {
               _NodeNameElement = "";
if(_NODE.equals(localName)){
               System.out.println("NODENAME: %%%%%%%%%");
please help me. How can I figure node ending for particular nodename

Error while encrypting the xml file using asymmetric cipher...pls help

i am encryption the xml file using asymmetric cyper....
in one class , i am generating private key and public key using "RSA" algorithm..using 1024 byte initialization...
it is generating properly...
and after that in second class i am encrypting the xml file using "DESede" algorithm and i am using public key of above class..
but i am getting exceptione :
java.security.InvalidKeyException: Invalid key length: 162 bytes
at com.sun.crypto.provider.DESedeCipher.engineGetKeySize(DashoA6275)
at javax.crypto.Cipher.init(DashoA6275)
at XmlEncryption.getEncryptedData(XmlEncryption.java:147)
at XmlEncryption.encryptCompleteXmlFile(XmlEncryption.java:123)
at demoXmlEncApp.simulateBookSellersEnd(demoXmlEncApp.java:72)
at demoXmlEncApp.main(demoXmlEncApp.java:29)
so, what is that ?
i want to use RSA algo for key generatiion and DESede for cipher initialization .
is there any site for where source code of xml file encryption using asymmetric cipher is available?
pls, help me.....

Sachin,
What sabre150 is trying to explain to you, is that encrypting data (such as an XML file) is a 2-step process. In the first step, you generate a symmetric key (such as DES, 3DES, AES) and encrypt your file with this key (and the appropriate mode and padding).
In order to ensure that your encrypted data cannot be decrypted by unauthorized individuals, you now need to protect the symmetric key. Encrypting your symmetric key with another symmetric key does not solve the problem, since you have a chain of symmetric keys that need to be encrypted into infinity.
Cryptographers solve this problem by encrypting the symmetric key with an asymmetric key, such as as RSA. To perform this second step, you generate an RSA key-pair, encrypt the symmetric key you generated in step 1 with the Public key of the RSA key-pair, and give the recipient of the encrypted XML file, access to the Private key of the RSA key-pair as well as the encrypted symmetric key.
The recipient, then uses the RSA Private key to decrypt the symmetric key first, and then uses the "plaintext" symmetric key to decrypt the XML file.
What sabre150 was also attempting to explain to you, is the traditional way of transporting an encrypted "blob" that consists of data + symmetric key. With a modern key-management system, combining the two would be unnecessary.

Problem in accessing XML files using Web Start

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