Transfer 100M XML file with XSL

Hi,
I am trying to transfer 100M XML file with XSL. Input.xml is the XML file, format.xsl is the XSL file. I type in the command line as:
java org.apache.xalan.xslt.Process -IN input.xml -XSL format.xsl -OUT output.xml
It got "out of memeory" error. My questions are:
1. Is it possible to transfer such large XML file with XSLT?
2. The XSL processor used SAX or DOM to parse XML file?
3. Any suggestions?
Thanks.
James

maybe?
java -Xmx200m org.apache.xalan.xslt.Process -IN input.xml -XSL format.xsl -OUT output.xml
http://java.sun.com/j2se/1.3/docs/tooldocs/win32/java-classic.html

Similar Messages

Split XML files with XSL result document

Hi All,
I have below xml file...
<?xml version="1.0" encoding="ISO-8859-1"?>
<T0020
xsi:schemaLocation="http://www.safersys.org/namespaces/T0020V1 T0020V1.xsd"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://www.safersys.org/namespaces/T0020V1">
<INTERFACE>
<NAME>SAFER</NAME>
<VERSION>04.02</VERSION>
</INTERFACE>
<TRANSACTION>
<VERSION>01.00</VERSION>
<OPERATION>REPLACE</OPERATION>
<DATE_TIME>2009-09-01T00:00:00</DATE_TIME>
<TZ>CT</TZ>
</TRANSACTION>
<IRP_ACCOUNT>
<IRP_CARRIER_ID_NUMBER>274845</IRP_CARRIER_ID_NUMBER>
<IRP_BASE_COUNTRY>US</IRP_BASE_COUNTRY>
<IRP_BASE_STATE>AR</IRP_BASE_STATE>
<IRP_ACCOUNT_NUMBER>55002</IRP_ACCOUNT_NUMBER>
<IRP_ACCOUNT_TYPE>I</IRP_ACCOUNT_TYPE>
<IRP_STATUS_CODE>100</IRP_STATUS_CODE>
<IRP_STATUS_DATE>2007-11-06</IRP_STATUS_DATE>
<IRP_UPDATE_DATE>2009-08-03</IRP_UPDATE_DATE>
<IRP_NAME>
<NAME_TYPE>LG</NAME_TYPE>
<NAME>A P SUPPLY CO</NAME>
<IRP_ADDRESS>
<ADDRESS_TYPE>PH</ADDRESS_TYPE>
<STREET_LINE_1>1400 N OATS</STREET_LINE_1>
<STREET_LINE_2/>
<CITY>TEXARKANA</CITY>
<STATE>AR</STATE>
<ZIP_CODE>71854</ZIP_CODE>
<COUNTY>MILLER</COUNTY>
<COLONIA/>
<COUNTRY>US</COUNTRY>
</IRP_ADDRESS>
<IRP_ADDRESS>
<ADDRESS_TYPE>MA</ADDRESS_TYPE>
<STREET_LINE_1>P O BOX 1927</STREET_LINE_1>
<STREET_LINE_2/>
<CITY>TEXARKANA</CITY>
<STATE>AR</STATE>
<ZIP_CODE>75504</ZIP_CODE>
<COUNTY/>
<COLONIA/>
<COUNTRY>US</COUNTRY>
</IRP_ADDRESS>
</IRP_NAME>
</IRP_ACCOUNT>
<IRP_ACCOUNT> ..... </IRP_ACCOUNT>
<IRP_ACCOUNT> ..... </IRP_ACCOUNT>
<IRP_ACCOUNT> ..... </IRP_ACCOUNT>
</T0020>
and i want to take this xml file and split it into multiple files through java code like this ...
File1.xml
<T0020>
<IRP_ACCOUNT> ..... </IRP_ACCOUNT>
<IRP_ACCOUNT> ..... </IRP_ACCOUNT>
</T0020>
File2.xml
<T0020>
<IRP_ACCOUNT> ..... </IRP_ACCOUNT>
<IRP_ACCOUNT> ..... </IRP_ACCOUNT>
</T0020>
so i have applied following xslt ...
<?xml version="1.0" encoding="UTF-8" ?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:t="http://www.safersys.org/namespaces/T0020V1" version="2.0">
<xsl:output method="xml" indent="yes" name="xml" />
<xsl:variable name="accounts" select="t:T0020/t:IRP_ACCOUNT" />
<xsl:variable name="size" select="10" />
<xsl:template match="/">
<xsl:for-each select="$accounts[position() mod $size = 1]">
<xsl:variable name="filename" select="resolve-uri(concat('output/',position(),'.xml'))" />
<xsl:result-document href="{$filename}" format="xml">
<T0020>
<xsl:for-each select=". | following-sibling::t:IRP_ACCOUNT[position() < $size]">
<xsl:copy-of select="." />
</xsl:for-each>
</T0020>
</xsl:result-document>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
Now i want to apply this XSL to xml through Java Code...
TransformerFactory tFactory = TransformerFactory.newInstance();
Source xslSource = new StreamSource(xslFilePath);
Transformer trans = tFactory.newTransformer(xslSource);
trans.transform(new StreamSource(xmlFileName), new StreamResult( ????));
here how can i map new StreamResult( ) input parameter with xsl Result document argument ??
Please help me.....
Or Can you give me a link of Example which use result document and Java transform method to Output multiple doucment ??
Here new StreamResult take only 1 file as to be transformed ....

hi Tejas ,
I have done as you said but now able to generate multiple xml file .
I am giving you the xsl file i have used ....
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:t="http://www.safersys.org/namespaces/T0020V1" version="2.0">
     <xsl:output method="xml" indent="yes" name="xml" />
     <xsl:variable name="accounts" select="t:T0020/t:IRP_ACCOUNT" />
     <xsl:variable name="size" select="5" />
     <xsl:template match="/">
          <xsl:for-each select="$accounts[position() mod $size = 1]">
               <xsl:variable name="filename" select="concat(position(),'.xml')" />
                    <xsl:result-document href="{$filename}" format="xml">
                         <T0020>
                              <xsl:for-each select=". | following-sibling::t:IRP_ACCOUNT[position() < $size]">
                                   <xsl:copy-of select="." />
                              </xsl:for-each>
                         </T0020>
                    </xsl:result-document>
          </xsl:for-each>
     </xsl:template>
</xsl:stylesheet>
and i have done transformation like this ...
Transformer trans = tFactory.newTransformer(xslSource);
trans.transform(new StreamSource(xmlFileName), new DOMResult());
but not getting any result .
Can you please help me out ?
- Nisarg

Transform Large XML files with XSL

HELP, LARGE XML FILES
I have got 30 - 50 MB large xml file, and i would like to transform it
with xslt, i tried but i have got OutOfMemory Exception.
I tried to find out the solution on JAVA site, but i didn't find it.
I can not displit my xml file. I hope for some help.
I tried really everything.
Thanks a lot

What is your machine configuration ?
The above 2 suggestions would help, but it does depend on how your software is written.
Please post more info about your environment and object design.
Chintan

Xml file with xsl style sheet reference

I am trying to learn this from a book, I copied the code like
it shows, but the page on the server keeps coming back with an
error starting on line 54 involving the four "End If" quotes. I
have tried everything I can think of and can't get it to work. Any
help would be greatly appreciated.
Here's the code:
<%@Language=VBScript%>
<%option explicit%>
<%
dim domDoc
dim aFilename
Dim node
Dim nodes
dim mainNodes
dim mainnode
dim attr
dim root
Set domDoc = Server.CreateObject("MSXML.DOMDocument")
afilename =server.MapPath("ch19i1.xml")
if not domDoc.load(afilename) then
Response.Write "Could not load the file " & aFilename
& "<br>"
Response.End
end if
%>
<html>
<head></head>
<body>
<%
' display the root node
set root = domDoc.documentElement
Response.Write "<h1>" & root.nodename & ": "
& root.nodevalue & "</h1>"
Set mainNodes = root.childNodes
For Each mainNode in mainNodes
If mainNode.hasChildNodes = False Then
Response.Write "<h2>" & mainNode.nodeName & ":
" & mainNode.text & "</h2><br>"
Else
Response.Write "<h3>" & mainNode.nodeName & ":
" & mainNode.text & "</h3>"
Set nodes = mainNode.childNodes
If nodes.lenght > 0 Then
Set node = nodes(0)
If Not node.Attributes Is Nothing Then
If node.Attributes.lenght > 0 Then
Response.Write "<table border='1'>"
Response.Write "<thead>"
For Each attr In node.Attributes
Response.Write "<th>" & attr.nodeName &
"</th>"
Next
Response.Write "</tr>"
For Each node In nodes
Response.Write "<tr>"
For Each attr In node.Attributes
If attr.nodeValue = "" Then
Response.Write "<td>" & " " &
"</td>"
Else
Response.Write "<td>" & attr.nodeValue &
"</td>"
End If
Next
Response.Write "</table>"
End If
End If
End IF
End If
Next
%>
</body>
</html>

Is the error message displayed in Firefox or in IE, or in a customized window that doesn't identify the browser?
''If it displays in Firefox:''
It's possible that the Troubleshooter doesn't work correctly unless IE is your default browser. You could test that possibility by having IE make itself the default and testing the Troubleshooter again.
''If it displays in IE or embedded in another Microsoft application:''
In a web search I found these suggestions:
(1) Reset your Internet Explorer settings, according to http://answers.microsoft.com/en-us/ie/forum/ie8-windows_7/cannot-view-xml-using-xsl-style-sheet/ccfe80c6-c0db-4594-a7e3-475f9eac0e85
(2) Try the System File Checker, according to http://ask-leo.com/why_do_i_get_the_xml_page_cannot_be_displayed_after_running_a_microsoft_troubleshooter.html
Any luck?

Issue with reading a xml file from xsl

Hi,
When I am trying to read a xml file from xsl, I am getting unwanted output.
Following is the XSL:
<?xml version="1.0" encoding="UTF-8" ?>
<?oracle-xsl-mapper

<mapSources>
    <source type="XSD">
      <schema location="../xsd/B2BMarketProperties.xsd"/>
      <rootElement name="ReceipentIDType" namespace="http://www.example.org"/>
    </source>
</mapSources>
<mapTargets>
    <target type="XSD">
      <schema location="../xsd/B2BMarketProperties.xsd"/>
      <rootElement name="ReceipentIDType" namespace="http://www.example.org"/>
    </target>
</mapTargets>

?>
<xsl:stylesheet version="1.0"
                xmlns:bpws="http://schemas.xmlsoap.org/ws/2003/03/business-process/"
                xmlns:xp20="http://www.oracle.com/XSL/Transform/java/oracle.tip.pc.services.functions.Xpath20"
                xmlns:mhdr="http://www.oracle.com/XSL/Transform/java/oracle.tip.mediator.service.common.functions.MediatorExtnFunction"
                xmlns:bpel="http://docs.oasis-open.org/wsbpel/2.0/process/executable"
                xmlns:oraext="http://www.oracle.com/XSL/Transform/java/oracle.tip.pc.services.functions.ExtFunc"
                xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
                xmlns:ns0="http://www.example.org"
                xmlns:dvm="http://www.oracle.com/XSL/Transform/java/oracle.tip.dvm.LookupValue"
                xmlns:hwf="http://xmlns.oracle.com/bpel/workflow/xpath"
                xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
                xmlns:med="http://schemas.oracle.com/mediator/xpath"
                xmlns:ids="http://xmlns.oracle.com/bpel/services/IdentityService/xpath"
                xmlns:bpm="http://xmlns.oracle.com/bpmn20/extensions"
                xmlns:xdk="http://schemas.oracle.com/bpel/extension/xpath/function/xdk"
                xmlns:xref="http://www.oracle.com/XSL/Transform/java/oracle.tip.xref.xpath.XRefXPathFunctions"
                xmlns:xsd="http://www.w3.org/2001/XMLSchema"
                xmlns:ora="http://schemas.oracle.com/xpath/extension"
                xmlns:socket="http://www.oracle.com/XSL/Transform/java/oracle.tip.adapter.socket.ProtocolTranslator"
                xmlns:ldap="http://schemas.oracle.com/xpath/extension/ldap"
                exclude-result-prefixes="xsi xsl ns0 xsd bpws xp20 mhdr bpel oraext dvm hwf med ids bpm xdk xref ora socket ldap">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:variable name="ReceipentID" select="document('../xsd/B2BMarketProperties.xml')"/>
<xsl:template match="/">
    <ns0:ReceipentIDType>
    <xsl:for-each select="$ReceipentID">
      <ns0:ReceipentID>
        <xsl:value-of select="$ReceipentID"/>
      </ns0:ReceipentID>
      </xsl:for-each>
    </ns0:ReceipentIDType>
</xsl:template>
</xsl:stylesheet>
Following is the XML ( B2BMarketProperties.xml)
<?xml version="1.0" encoding="UTF-8" ?>
<ReceipentIDType xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
                 xsi:schemaLocation="http://www.example.org B2BMarketProperties.xsd"
                 xmlns="http://www.example.org">
<ReceipentID>123</ReceipentID>
<ReceipentID>345</ReceipentID>
</ReceipentIDType>
The output i am getting with this code is
<?xml version="1.0" encoding="UTF-8"?>
<ns0:ReceipentIDType xmlns:ns0="http://www.example.org">
    <ns0:ReceipentID>123345</ns0:ReceipentID>
</ns0:ReceipentIDType>
But, I need output in the following format
<ns0:ReceipentIDType xmlns:ns0="http://www.example.org">
    <ns0:ReceipentID>123</ns0:ReceipentID>
     <ns0:ReceipentID>345</ns0:ReceipentID>
</ns0:ReceipentIDType>
Could you guys let me know what i am doing wrong. Any help would be appreciated.
Thanks,

This worked for me :
<xsl:template match="/">
    <ns0:ReceipentIDType>
      <xsl:for-each select="document('B2BMarketProperties.xml')/*:ReceipentIDType/*:ReceipentID">
        <xsl:variable name="count" select="position()"/>
        <ns0:ReceipentID>
          <xsl:value-of select="document('B2BMarketProperties.xml')/*:ReceipentIDType/*:ReceipentID[$count]"/>
        </ns0:ReceipentID>
      </xsl:for-each>
    </ns0:ReceipentIDType>
</xsl:template>

Xml file with embedded xsl reports unknow error: An unknown error has occurred (805303f4)

Hi,
I have an xml file with embedded xsl link like below :
XML :
<pre><nowiki><?xml version="1.0" encoding="ISO-8859-1"?>
<?xml-stylesheet type="text/xsl" href="http://abc.com/x.xsl"?>
<catalog>
<cd>
<title>Empire Burlesque</title>
<artist>Bob Dylan</artist>
<country>USA</country>
<company>Columbia</company>
<price>10.90</price>
<year>1985</year>
</cd>
</catalog></nowiki></pre>
When opening xml file on IE, opera, safary ... it works ok. But when opening it on FireFox, i have an error like :
Error loading stylesheet: An unknown error has occurred (805303f4)
http://abc.com/x.xsl
How could i solve this problem ?

A good place to ask questions and advice about web development is at the mozillaZine Web Development/Standards Evangelism forum.
The helpers at that forum are more knowledgeable about web development issues.
You need to register at the mozillaZine forum site in order to post at that forum.
See http://forums.mozillazine.org/viewforum.php?f=25

How to transfer a xml file?

Dear friends,
I have a question about how to transfer a xml file between server and client using axis or weblogic.
My application needs to send xml files instead of primitive data types and bean objects.
Is it possible to tranfer files between server and client?
thank you very much for your great help!
jayanandan.

hi,
There are two ways in which u can send an xml file.
1) U send it along with the body of the soap message.
2)u can send it as an attachment in the soap envelope. this method u can use for any type of files.
If u send it in soap body the it will take more time for the processing of the soap body.so attachment is the best way.
I case of any queries please feel free to post it in the forum
Regards
Sandy

XML file with an attached MIME encoded ZIP file

Hi all,
I'm new to SAP WAS and MIME encoding/decoding, and I'm trying to generate an XML file with an attachment which is also MIME encoded.
1) I have dummy files (1.jpg, 2.jpg) and I'm trying to zip these files into one zip file (files.zip).
2) I'm trying to MIME encode/decode this zip file.
3) I'm trying to attach this MIME encoded zip file to existing XML file.
Which FMs could I use to accomplish this? Your help is very appreciated.
Thank you.
below is a file example that I'm trying to generate.
MIME-Version: 1.0
Content-Type: multipart/mixed;
     boundary="--XXXXboundary text"
Content-Transfer-Encoding: 7bit
This is a multi-part message in MIME format.
--XXXXboundary text
Content-Type: text/xml;
     charset="utf-8"
Content-Transfer-Encoding: 8bit
<?xml version="1.0" encoding="utf-8"?>
<abc/>
<def/>
--XXXXboundary text
Content-Type: application/octet-stream;      name="files.zip"
Content-Transfer-Encoding: base64
Content-Disposition: attachment
UEsDBBQAAAAIAI9EejAs5k34H84DAAYgBAAMAAAAMDAyMjQ5MTEucGRmnLsJWBNJ2zb6jmJIIIFE
BAMIJGEVBSKGRRAhEGQNoGwKYoiAEnYRUFGIkBAQFzYXRNHAAEGQZRy2ATFDUAHfGScSIUwQMMrM
ECGADptA0n/jzLtc//dd51znVAKdru6uqn7q6fu5764qQz

Just create an applet (extend JApplet)... add a JTextArea to it....fill the text area with the text from an XML doc.
To get the text of the XML doc just do something like..
File xmlFile = new File("<path to xml file>");
FileInputStream fis = new FileInputStream(xmlFile);
byte[] bytes = new byte[(int) xmlFile.length()];
fis.read(bytes);
fis.close();
String xmlText = new String(xmlBytes);
textArea.setText(xmlText);
...try something like that (assuming..i understand what it is u want)

Combine several XML files with same structure

Hello,
I have several XML files with the same structure and I want to combine them and create a new XML file to be bale to compare that information easily. It does not look very difficult but as I am very new in this I am not bale to get it
The structure of my actual files would be something simlar to:
Root->...-> Name->Address, Telephone
And what I would like to have si something like
Root->.... ->Address-> Name 1,Name 2....
Root -> ...->Telephone-> Name 1, Name 2....
Does anyone know how to do this.
Thanks

You could write a XSL transformation file that does this and transform your input file via
     * Transform XML file with a style sheet.
     * <p><b>Example:</b><p>
<table align="center" bgcolor="#E0E0E0" border=1 cellpadding="10" cellspacing="0"><tr><td><pre style="margin-top:0; margin-bottom:0">
XMLTransformer t = new XMLTransformer();
FileOutputStream fos = new FileOutputStream("C:/Project/result.html");
String xmlFile = "C:/Project/source.xml";
String styleSheet = "C:/Project/stylesheet.xsl";
t.transform(xmlFile, styleSheet, fos);
</pre></td></tr></table>
     * @param xmlfile The XML file to transform.
     * @param style Stylesheet to use for transformation.
     * @param outputStream OutputStream to write the transformed result to.
    public void transform(String xmlfile, String style, OutputStream outputStream) {
        DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
        //factory.setNamespaceAware(true);
        //factory.setValidating(true);
            //todo: use inputstreams instead of file names
            try {
                File stylesheet = new File(style);
                File datafile   = new File(xmlfile);
                DocumentBuilder builder = factory.newDocumentBuilder();
                document = builder.parse(datafile);
                // Use a Transformer for output
            TransformerFactory tFactory = TransformerFactory.newInstance();
            StreamSource stylesource = new StreamSource(stylesheet);
            Transformer transformer = tFactory.newTransformer(stylesource);
            DOMSource source = new DOMSource(document);
            StreamResult sr = new StreamResult(outputStream);
            transformer.transform(source, sr);
        } catch (TransformerConfigurationException tce) {
           // Error generated by the parser
           System.out.println ("\n**XMLTransformerr Factory error");
           System.out.println("   " + tce.getMessage() );
           // Use the contained exception, if any
           Throwable x = tce;
           if (tce.getException() != null) {
               x = tce.getException();
           x.printStackTrace();
        } catch (TransformerException te) {
           // Error generated by the parser
           System.out.println ("\n** Transformation error");
           System.out.println("   " + te.getMessage() );
           // Use the contained exception, if any
           Throwable x = te;
           if (te.getException() != null) {
               x = te.getException();
           x.printStackTrace();
         } catch (SAXException sxe) {
           // Error generated by this application
           // (or a parser-initialization error)
           Exception x = sxe;
           if (sxe.getException() != null) {
               x = sxe.getException();
           x.printStackTrace();
        } catch (ParserConfigurationException pce) {
            // Parser with specified options can't be built
            pce.printStackTrace();
        } catch (IOException ioe) {
           // I/O error
           ioe.printStackTrace();
    }//transform()

Loading xml file with multiple rows

I am loading data from xml files using xsl for transformation. I have created xsl's and loaded some of the data. In an xml file with multiple row, it's only loading one (the first) row. Any idea how I can get it to read and load all the records in the file???

Could some please help me with the above. I desparately need to move forward.

How to convert xml file to xsl using java

Hi all,
I have an XML file with which i need to convert(transform) it to an xsl file using java.
I am new to converting xml file to xslt.Please send me if u have code .
Thanks in advance
regards
Ram

You seem to be asking the wrong question. An XSL file can be used to transform an XML file, but transforming an XML into an XSL does not make sense. The API for running XSL transforms on the Java platform is described here:
http://java.sun.com/javase/6/docs/api/javax/xml/transform/package-summary.html

XML file with different format

Hello
   I have a requirement where the customer needs XML file with different format. How do I achieve the second format mentioned below instead of first format.
Normally we create the XML file in the following format:
- <Entries>
    - <Organization>
        <MemberName>0000000002</MemberName>
        <FullName>BAY GYNECOLOGICAL ASSOCIATES</FullName>
        <OrgId>0000000002</OrgId>
        <OrgType>CUST</OrgType>
   - <AssociatedToOrg>
        <Name>0002</Name>
   - </AssociatedToOrg>
</Organization>
Now the customer wants like the following format:
<Organization MemberName="522_Community_Customer" FullName="522 Community Customer" OrgId="DEA123456" OrgIdType="DEA" >
      <AssociatedToOrg Name="MN"/>
    </Organization>
Thanks
Naga

you can solve it with a XSL mapping. refer to the SDN to know how develop it
/people/aleksey.popov2/blog/2010/01/26/consuming-webservices-with-tag-in-wsdl-using-xslt
XSLT Editor for creating xlst mappings
http://www.sdn.sap.com/irj/scn/index?rid=/library/uuid/190eb190-0201-0010-0ab3-e69f70b6c257
http://wiki.sdn.sap.com/wiki/display/XI/XSLTMappingSteps
http://wiki.sdn.sap.com/wiki/display/XI/FileTOFile-UsingXSLTmapping%28forBeginners%29
http://wiki.sdn.sap.com/wiki/display/Snippets/ConvertflatXMLfiletonestedusing+XSLT
http://wiki.sdn.sap.com/wiki/display/Java/MultiMappingwithJavaandXSLTmappings
Edited by: Rodrigo Alejandro Pertierra on Oct 1, 2010 5:26 PM

How can I save a XML file with JAXP1.1?

Dear All.
I write a program to create XML file with DOM model, but I can't know how to save it? My environment is JAXP1.1 and JDK1.3.1,I has been required not use other XML parser toolkits,only JAXP1.1.
How can I do? thank you.
Many person give me a idea the com.sun.xml.tree.XmlDocument, but I can't find the class in API document or JAXP1.1's packages. why?
what is it? How can i use it?
thank you very much.

The way to save an XML Document is using a Transformer.
To have access to a transformer use the packages :
import javax.xml.transform.*;
import javax.xml.transform.dom.*;
import javax.xml.transform.stream.*;
Then for saving your Document Object (named dXml) get a Transformer Object with the TransformerFactory Object :
TransformerFactory tf = TransformerFactory.newInstance();
Transformer t = tf.newTransformer();
Now you have got your Transformer Object, to save your Document Object use the method :
Document dXml = getMyDocument(); // this is your Document Object.
OutputStream osSave = getMySaveStream(); // this the OutputStream you need to save your Document.
try
t.transform(new DomSource(dXml), new StreamResult(new OutputStreamWriter(osSave)));
finally
osSave.close();
And your Document was now saved.

Steps in converting a xml file with an rtf template to a pdf

Hey all,
What are the steps in converting a xml file with an rtf template to a pdf using XML Publisher from command line.
Thanks
Ravi

I don't have any code to do exactly what you wish, but it shouldn't be too difficult and http://www.dadhi.com/2007/06/generate-and-store-pdf-file-in-same.html is a good starting point.
Paul

Generation of Xml file with java output

Hi i m new to xml and java combination. I have a name value pair kind of output returning from java program. I want to generate the new xml file with the data. Could some one help me out in generating xml file with the data. Could anyone send me the java code that does this task.

Let me know which parser are you using currently for reading xml files so that i assist you. For now, you can refer to STAX Parser API under this link
http://java.sun.com/webservices/docs/1.6/tutorial/doc/SJSXP3.html

Transfer 100M XML file with XSL

Similar Messages

Maybe you are looking for