Want to view the created zip file via Bursting in Oracle Apps

Hello,
I' ve created a Bursting control file and invoked the concurrent request (XML Publisher Report Bursting Program, XDOBURSTREP) within the after-report trigger in oracle reports.
My Bursting-Control file as follows:
<xapi:requestset xmlns:xapi="http://xmlns.oracle.com/oxp/xapi" type="bursting">
<xapi:request select="/XXMI_OM_QUOTE/LIST_G_ORDERS/G_ORDERS">
<xapi:delivery>
<xapi:filesystem id="file1" output="/usr/tmp/DEV1/${QUOTE_NO}"></xapi:filesystem>
</xapi:delivery>
<xapi:document output-type="pdf" delivery="123">
<xapi:template type="xsl-fo" location="xdo://XXMI.XXMI_OM_QUOTE.en.US">
</xapi:template>
</xapi:document>
</xapi:request>
</xapi:requestset>
It works fine. e.g. Bursting creates two pdf-files and one zip file in unix, which I can see when pushing the View Output of the request(XML Publisher Report Bursting Program):
<?xml version="1.0" encoding="UTF-8" ?>
- <BURS_REPORT>
<REQUESTID>643527</REQUESTID>
<PARENT_REQUESTID>643525</PARENT_REQUESTID>
<REPORT_DESC>XXMI OM Quote</REPORT_DESC>
<OUTPUT_FILE>/u00/erp/dev1APP/inst/apps/DEV1/logs/appl/conc/out/o643527.zip</OUTPUT_FILE>
- <DOCUMENT_STATUS>
<KEY />
<OUTPUT_TYPE>pdf</OUTPUT_TYPE>
<DELIVERY>FILESYSTEM</DELIVERY>
<OUTPUT>/usr/tmp/DEV1/10000012.pdf</OUTPUT>
<STATUS>success</STATUS>
<LOG />
</DOCUMENT_STATUS>
- <DOCUMENT_STATUS>
<KEY />
<OUTPUT_TYPE>pdf</OUTPUT_TYPE>
<DELIVERY>FILESYSTEM</DELIVERY>
<OUTPUT>/usr/tmp/DEV1/10000026.pdf</OUTPUT>
<STATUS>success</STATUS>
<LOG />
</DOCUMENT_STATUS>
</BURS_REPORT>
But instead of this XML-output I wish a link to the file on unix. How can I link <OUTPUT_FILE>/u00/erp/dev1APP/inst/apps/DEV1/logs/appl/conc/out/o643527.zip</OUTPUT_FILE> within the view Output-Button of this request?
Thanks in advance for any clues

That is not how bursting works in EBS.
The View Output from XDOBURSTREP is the Bursting Status Reprot.
The bursted results are (potentially) multiple files so cannot be linked to view output.
Depending upon the delivery channel chosen you have to either read the email, check the fiel system or printer, to see the results.
The delivery channel bypasses the standard EBS request view.
Kevin

Similar Messages

How to displays the standard open file dialog box in Oracle Apps Server

Hi,
I am having a form, where i am opening a file and putting that file content in a text box.
This is my code written in WHEN-BUTTON-PRESSED trigger:
declare
in_file Text_IO.File_Type;
linebuf VARCHAR2(8000);
filename VARCHAR2(30);
BEGIN
filename:=GET_FILE_NAME('d:\Rajesh\Practice', File_Filter=>'Text Files (*.txt)|*.txt|',select_file=>false);
in_file := Text_IO.Fopen(filename, 'r');
LOOP
Text_IO.Get_Line(in_file, linebuf);
:text_item5:=substr(:text_item5||linebuf||chr(10),1,5000);
END LOOP;
EXCEPTION
WHEN no_data_found THEN
Text_IO.Put_Line('Closing the file...');
Text_IO.Fclose(in_file);
END;
I have posted this in ORacle Apps Server.Where i was not able to get the Open Dialoq Box.It is giving an error as
"FRM-40735:WHEN-BUTTON-PRESSED trigger raised unhandled exception ORA-302000.".
I think the GET_FILE_NAME function is not supported in the Apps Server.
Is there any alternative solution for getting the OPEN DIALOG WINDOW in Oracle Apps Server?
Pls let me know.

Hi!
waldemar.hersacher wrote:
"It seems that the VIs called by the clients are running in the user interface thread.
A call to the file dialog box will call a modal system dialog. So LV can't go on executing VIs in the user interface thread."
Are you sure? I think, that File Dialog, called by LabVIEW File Dialog from Advanced File Functions Palette doesn't blocking any cycles in subVI, which running in UI Thread. Look in my old example (that was prepared for other topic, but may be good for this topic too).
2 linus:
I think, you must a little bit reorganize you application for to do this. You must put your File Dialog into separated cycle. This cycle, of course, will be blocked when File Dialog appear on the screen. But other cy
cle, which responsible for visualization will run continuosly at the same time...
Attachments:
no_block_with_file_open_dialog.zip ‏42 KB

Want To create Zip file using java,And Unzip without Java Program

I want to create a zip text file using java, I know Using ZipOutputStream , we can create a zip file, , But i want to open that zip file without java program. suppose i use ZipOutputStream , then zip file is created But for unZip also difftrent program require. We cant open that zip file without writing diff java program.
Actually i have one text file of big size want to create zip file using java , and unzip simply without java program.Its Possible??
Here is one answer But I want to open that file normal way(
For Exp. using winzip we can create a zip file and also open simply)
http://forum.java.sun.com/thread.jspa?threadID=5182691&tstart=0

Thanks for your Reply,
I m creating a zip file using this program, Zip file Created successfully But when im trying to open .zip file i m getting error like "Canot open a zip file, it does not appear to be valid Archive"
import java.io.*;
import java.util.zip.*;
public class ZipFileCreation
     public static void main (String argv[])
     try {
     FileOutputStream fos = new FileOutputStream ( "c:/a.zip" );
     ZipOutputStream zip = new ZipOutputStream ( fos );
     zip.setLevel( 9 );
     zip.setMethod( ZipOutputStream.DEFLATED );
//     get the element file we are going to add, using slashes in name.
     String elementName = "c:/kalpesh/GetSigRoleInfo092702828.txt";
     File elementFile = new File ( elementName );
//     create the entry
     ZipEntry entry = new ZipEntry( elementName );
     entry.setTime( elementFile.lastModified() );
//     read contents of file we are going to put in the zip
     int fileLength = (int)elementFile.length();
     System.out.println("fileLength = " +fileLength);
     FileInputStream fis = new FileInputStream ( elementFile );
     byte[] wholeFile = new byte [fileLength];
     int bytesRead = fis.read( wholeFile , 0 /* offset */ , fileLength );
//     checking bytesRead not shown.
     fis.close();
//     no need to setCRC, or setSize as they are computed automatically.
     zip.putNextEntry( entry );
//     write the contents into the zip element
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//     close the entire zip
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e.printStackTrace();
}

PI needs to obtain a zip file via FTP using the File adapter

I have a scenario where PI needs to obtain a zip file via FTP using the File adapter, this zip file contains a number of txt files that I need to process, and the content of one of them send it to an ECC, now I'm using the PayloadZipBean Module in the Sender FIle Adapter, and I have two things if I use the Message Protocol as File, I get a Payload for each txt file in the zip file, but this payload has no structure, and if I use the File Content Conversion I get an XML strcuture with only one field and a strange string in it, and somewhere in this string the names of the files I assume all the content of the zip file, can anyone help on how could I achieve what I need that is to pull the zip file via SAP PI, then unzip it, and with the content of one of the txt files send it to an ECC via ABAP Proxy, thanks in advance for your answers.
Regards,
Raul Alvarado

Hello Raul,
you can do it in futher way ...
pickup zip file and simply extract and dump it in another temp folder (can use scripts on OS level).
@ then Use another sender communication channel to pickup all these text file .
for further clarification you can use these links also. -
Process txt files in zip file
Accessing File using FTP from Java Mapping
File Sender Adapter with FTP protocol
BR
Raj

Hi I am using Iframe with src="abc.mht" file. i want to view the mht file content in same window.please help.. Thanks in Advance

Hi I am using Iframe with src="abc.mht" file. i want to view the mht file content in same window.please help..
Thanks in Advance

This is not a Java problem, this is a WEBBOT problem.
I suggest that you research the topic in your FrontPage help files.
Yes, we do know how to solve this problem using ordinary HTML and JSP. The sample code that you've posted implies that you are looking for an extremely different solution. To put it another way, it looks like you're logging onto the Federal Aviation Administration's forums and asking them how to go from Chicago to New York by train.
But in any case, if you can't find the information in FrontPage, I'm 100% certain that if you went to www.microsoft.com and searched for FrontPage and downloads, and skipped the results that actually want you to download FrontPage, you'll get some sample code.

From where can be the WebCenterSRDemo.zip file downloaded ?

Hi,
Oracle® WebCenter Framework Developer's Guide
"2.3.2 Setting Up the Oracle ADF Service Request Demo
To view the SRDemo and perform some of the example tasks in this guide, you must download and install the starter files. To do so, perform the following steps:
Download the WebCenterSRDemo.zip file, located on this page:
http://www.oracle.com/technology/products/webcenter/documentation.html
This is not the right link and the WebCenterSRDemo is not available via JDeveloper "Check for Updates Wizard".
Please, inform me of its location, if existing...
Regards
Boris

Thank you Peter,
can you inform us also about the integration with the Oracle SOA Suite :
3.3 Enabling Oracle SOA Suite or a Standalone OC4J for WebCenter Applications
" If you have Oracle SOA Suite 10.1.3.1.0 or a standalone OC4J 10.1.3.1.0, then you can update it to create, deploy, and run Oracle WebCenter Suite 10.1.3.2.0 applications. Patches to enable you to run WebCenter applications will be released in the future. Watch for more information about MetaLink and the Oracle Technology Network:
http://www.oracle.com/technology/products/webcenter/index.html..."
What I'm missing in the WebCenter Suite is a tool for Business Process Modelling , like Business Process Architect (ARIS ToolSet)...
Regards
Boris

Where to download the webcentertutorialcontent.zip file for JDev 10.1.3.x?

A quick question for the download location of the webcentertutorialcontent.zip file for version 10g (10.1.3.2 or later).
I am going through the Oracle WebCenter Framework Tutorial 10g (10.1.3.2.0) (document id B31072-02) to gain an understanding of the working mechanism of portlets. The tutorial requires the use of the webcentertutorialcontent.zip file to set up the sample application, and points to http://www.oracle.com/technology/products/webcenter/files/webcentertutorialcontent.zip for downloading the file.
But that page has been updated to 11g. I was able to find the archived old versions of webcenter, but not that zip file.
It would very much appreciated if someone knows and let me know where it is.
Many thanks!
Newman

On this page: http://www.oracle.com/technetwork/middleware/webcenter/documentation/index.html you have a list of all files.
At the bottom in the Oracle WebCenter 10g (10.1.3.x) section you have a part: Building a WebCenter Application Step by Step Guide with a link to the supported files: http://download.oracle.com/otndocs/tech/webcenter/files/SRDemo_App_Download.zip
This is from the 10g version...
You are right that 11g applications does not run on a 10g app server. Oracle 11g fusion middelware uses weblogic as an application server. This does not use the OC4J container anymore.
11g and 10g have a different licence so if you are using 10g and you want to use 11g, you need to buy a licence for weblogic and so on.
If you are currently looking into the use of webcenter/portlets and you are planning to use it later on, than oracle will propose 11g because i don't think 10g is supported... I don't think you can buy a webcenter 10g license..
By the way... You can install jdeveloper 11g in a sepperate middelware folder than your 10g version. This means that you easily can use them together. If you are planning on using webcenter, the best thing to do is install jdeveloper 11.1.1.3 (latest stable version) and install the webcenter extension. This allow you to easily create portlet producers and webcenter applications. You can easily deploy them on the integrated server. This is a lot easier than in the 10g release.

Create Zip File In Windows and Extract Zip File In Linux

I had created a zip file (together with directory) under Windows as follow (Code are picked from [http://www.exampledepot.com/egs/java.util.zip/CreateZip.html|http://www.exampledepot.com/egs/java.util.zip/CreateZip.html] ) :
package sandbox;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.zip.ZipEntry;
import java.util.zip.ZipOutputStream;
* @author yan-cheng.cheok
public class Main {
     * @param args the command line arguments
    public static void main(String[] args) {
        // These are the files to include in the ZIP file
        String[] filenames = new String[]{"MyDirectory" + File.separator + "MyFile.txt"};
        // Create a buffer for reading the files
        byte[] buf = new byte[1024];
        try {
            // Create the ZIP file
            String outFilename = "outfile.zip";
            ZipOutputStream out = new ZipOutputStream(new FileOutputStream(outFilename));
            // Compress the files
            for (int i=0; i<filenames.length; i++) {
                FileInputStream in = new FileInputStream(filenames);
// Add ZIP entry to output stream.
out.putNextEntry(new ZipEntry(filenames[i]));
// Transfer bytes from the file to the ZIP file
int len;
while ((len = in.read(buf)) > 0) {
out.write(buf, 0, len);
// Complete the entry
out.closeEntry();
in.close();
// Complete the ZIP file
out.close();
} catch (IOException e) {
e.printStackTrace();
The newly created zip file can be extracted without problem under Windows, by using [http://www.exampledepot.com/egs/java.util.zip/GetZip.html|http://www.exampledepot.com/egs/java.util.zip/GetZip.html]
However, I realize if I extract the newly created zip file under Linux, using modified version of [http://www.exampledepot.com/egs/java.util.zip/GetZip.html|http://www.exampledepot.com/egs/java.util.zip/GetZip.html] . The original version doesn't check for directory using zipEntry.isDirectory()).public static boolean extractZipFile(File zipFilePath, boolean overwrite) {
InputStream inputStream = null;
ZipInputStream zipInputStream = null;
boolean status = true;
try {
inputStream = new FileInputStream(zipFilePath);
zipInputStream = new ZipInputStream(inputStream);
final byte[] data = new byte[1024];
while (true) {
ZipEntry zipEntry = null;
FileOutputStream outputStream = null;
try {
zipEntry = zipInputStream.getNextEntry();
if (zipEntry == null) break;
final String destination = Utils.getUserDataDirectory() + zipEntry.getName();
if (overwrite == false) {
if (Utils.isFileOrDirectoryExist(destination)) continue;
if (zipEntry.isDirectory())
Utils.createCompleteDirectoryHierarchyIfDoesNotExist(destination);
else
final File file = new File(destination);
// Ensure directory is there before we write the file.
Utils.createCompleteDirectoryHierarchyIfDoesNotExist(file.getParentFile());
int size = zipInputStream.read(data);
if (size > 0) {
outputStream = new FileOutputStream(destination);
do {
outputStream.write(data, 0, size);
size = zipInputStream.read(data);
} while(size >= 0);
catch (IOException exp) {
log.error(null, exp);
status = false;
break;
finally {
if (outputStream != null) {
try {
outputStream.close();
catch (IOException exp) {
log.error(null, exp);
break;
if (zipInputStream != null) {
try {
zipInputStream.closeEntry();
catch (IOException exp) {
log.error(null, exp);
break;
} // while(true)
catch (IOException exp) {
log.error(null, exp);
status = false;
finally {
if (zipInputStream != null) {
try {
zipInputStream.close();
} catch (IOException ex) {
log.error(null, ex);
if (inputStream != null) {
try {
inputStream.close();
} catch (IOException ex) {
log.error(null, ex);
return status;
*"MyDirectory\MyFile.txt" instead of MyFile.txt being placed under folder MyDirectory.*
I try to solve the problem by changing the zip file creation code to
+String[] filenames = new String[]{"MyDirectory" + "/" + "MyFile.txt"};+
But, is this an eligible solution, by hard-coded the seperator? Will it work under Mac OS? (I do not have a Mac to try out)
p/s To be honest, I do a cross post at [http://stackoverflow.com/questions/2549766/create-zip-file-in-windows-and-extract-zip-file-in-linux|http://stackoverflow.com/questions/2549766/create-zip-file-in-windows-and-extract-zip-file-in-linux] Just want to get more opinion on this.
Edited by: yccheok on Apr 26, 2010 11:41 AM

Your solution lies in the File.separator constant; this constant will contain the path separator that is used by the operating system. No need to hardcode one, Java already has it.
edit: when it comes to paths by the way, I have the bad habit of always using the front slash ( / ). This will also work under Windows and has the added benefit of not needing to be escaped.

Can't import the exercise zip files

I downloaded the exercise zip files and tried to import them to flex 3.
I can browse to the zip but flex will not let me finish.

Not sure what you are saying.
Downlaoded the exercises to my HD. Created a folder "Flex in a week" and tried to import the exercises.
Same problem wirh some books. Download the source code as a large zip. unzip the file and try to import projeect folder or project zip and can't.
Must be operator error software can't be that hard
Herman

Installing JDK 1.4, esp. the src.zip file

I have some questions about installing JDK 1.4:
1. After running the executable installation file, you get a src.zip file in the installation directory. Does anyone know what this is? The obvious answer is that it is the library of code for JDK 1.4. However, I have run programs on Forte Agent without unzipping this file.
2. In previous versions of the JDK, when you decompress the src.jar (it is now a src.zip file in the 1.4 version) file, it creates a src folder and decompresses all the information into that. However, in JDK 1.4, it does not create a src folder, but rather, decompresses the files into the installation directory. Is this a fault of the new release or should I create a src folder and have the src.zip file decompress into that folder?
I am running Windows 2000.

I have some questions about installing JDK 1.4:
1. After running the executable installation file, you
get a src.zip file in the installation directory. Does
anyone know what this is? The obvious answer is that
it is the library of code for JDK 1.4. However, I have
run programs on Forte Agent without unzipping this
file.The src.zip is the source code for the JDK, not library. Note: The file is not the whole source code, it does not have any native library so you cannot compile it.
>
2. In previous versions of the JDK, when you
decompress the src.jar (it is now a src.zip file in
the 1.4 version) file, it creates a src folder and
decompresses all the information into that. However,
in JDK 1.4, it does not create a src folder, but
rather, decompresses the files into the installation
directory. Is this a fault of the new release or
should I create a src folder and have the src.zip file
decompress into that folder?
I am running Windows 2000.No, it is just how Sun zip it. Anyway, as long as you don't care to look at the source, you can just leave it along.

Getting IO exception when trying to view the joblog.log file

Hi,
We are unable to view the joblog.log file. The joblog file is created in server path, but when we trying to view/read/download the joblog.log file we are getting JCS-108004: An IO exception occurred accessing a local file error.
server path: xxx/log/processerver_.log.
But we are able to view the reports through redwood for these extensions (*. html, *.csv. *. xml.).
server path: xxx/lis/sys_.html
Can anybody help on this issue.
Thanks,
Suresh.

Dear Suresh,
Your Question is not clear to me. what i understood is you can abel to check log in redwood , and unable to open in server at os level? if it is correct, login to server with cps installation user, ten i hope you can able to open files, check the file permissions in init file in $JCS_HOME/admin directory, check the permissions given, and also check listener port value, is it correct?
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Error when creating ZIP file (return value 2 when IGS was called)

Hello All
I attempted to Schedule a Query to the Portal folders and executes fine. However when it's scheduled to and email, I am getting this message
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Any ideas?
Thanks ...BK

Hello Kai
Thanks very much for the info. how would I know what patch the IGS is?
Thank you.
Regards..BK

The src.zip file

hi guys
I'm rather a beginner to Java.
I posted a question related to this one on another section.
I'd like to know about the src.zip file
Is it essential to set up your JRE_SRC variable to point to the src.zip file?
What is the purpose of src.zip?
If we want to reference types such as String or Boolean or Integer, do we need to have src.zip set up?
Thanks very much

Shane7070 wrote:
I'm not sure what JRE_SRC does as i've seen very limited documentation but from my understanding it's the java runtime source and is supposed to contain the src.zip file. That's what i gather from the documentationAre you talking about some folder/directory? As I said I have never heard of it.
flounder, is src.zip necessary in order to reference classes in the packages that it contains?
In one place I read that it is necessary but in another place I read that it's just the source code to the classes that are compiled and located elsewhere in the jdk.As I said the src.zip file contains the source code, that is the .java files. To run Java programs you need the .class files and these have been placed inside a .jar file.

Creating Zip File

I created a zip file by clicking on "Creative Archive," but the file size is still the same. Am I doing something wrong or is there another way to shrink the file size?

Hi astinn,
What type of file you're trying to compress and what type of compression program you're using makes a lot of difference.
As a quick test, I zipped a handful of PDF files on my drive. The resulting .zip file was indeed not much different than the total of the originals. What the PDF contains would also make a difference. If it's mostly just text, than the PDF is already highly compressed. If it has images in it, then it depends on if they're high res (300 dpi) press ready images, which will compress quite a bit, or already highly compressed embedded JPEGs.
Zip is very good at some things, not very good at others. Likewise with Stuffit. It can't compress things like Windows .wmf files at all. Whereas RAR will cut the size of that same file by about half.

Creating zip files in pl/sql.

I'm using code posted on this forum to create zip file in pl/sql. Works greate for combing multiple files into one archive. That's until I have to add a zip file.
Here's my test:
drop table t1;
create table t1 (file_name varchar2(100), file_blob blob);
declare
v_new_blob blob;
v_file_name varchar2(100);
v_buffer_raw raw(1000);
v_length integer;
b_zipped_blob BLOB;
zip_files zz_zip.file_list;
t_file blob;
begin
     dbms_lob.createtemporary(v_new_blob,true);
     v_buffer_raw := UTL_RAW.cast_to_raw('AAAA');
     v_length := UTL_RAW.length(v_buffer_raw);
     dbms_lob.writeappend(v_new_blob, v_length, v_buffer_raw);
     insert into t1 values ('tmp\FileA.txt',v_new_blob);
     dbms_lob.freetemporary(v_new_blob);
     dbms_lob.createtemporary(v_new_blob,true);
     v_buffer_raw := UTL_RAW.cast_to_raw('BBBBBBBBB');
     v_length := UTL_RAW.length(v_buffer_raw);
     dbms_lob.writeappend(v_new_blob, v_length, v_buffer_raw);
     insert into t1 values ('tmp\FileB.txt',v_new_blob);
     dbms_lob.freetemporary(v_new_blob);
     dbms_lob.createtemporary(v_new_blob,true);
     v_buffer_raw := UTL_RAW.cast_to_raw('CCCCCCCCCCCCC');
     v_length := UTL_RAW.length(v_buffer_raw);
     dbms_lob.writeappend(v_new_blob, v_length, v_buffer_raw);
     insert into t1 values ('tmp\FileC.txt',v_new_blob);
     dbms_lob.freetemporary(v_new_blob);
     dbms_lob.createtemporary(v_new_blob,true);
     v_buffer_raw := UTL_RAW.cast_to_raw('DDDDDDDDDDDDDDDDDDDDDDDDDD');
     v_length := UTL_RAW.length(v_buffer_raw);
     dbms_lob.writeappend(v_new_blob, v_length, v_buffer_raw);
     insert into t1 values ('tmp\FileD.txt',v_new_blob);
     dbms_lob.freetemporary(v_new_blob);
     commit;
     select file_name, file_blob into v_file_name, v_new_blob
          from t1 where file_name = 'tmp\FileA.txt';
     zz_zip.add1file(b_zipped_blob, v_file_name, v_new_blob);
     select file_name, file_blob into v_file_name, v_new_blob
          from t1 where file_name = 'tmp\FileB.txt';
     zz_zip.add1file(b_zipped_blob, v_file_name, v_new_blob);
     select file_name, file_blob into v_file_name, v_new_blob
          from t1 where file_name = 'tmp\FileC.txt';
     zz_zip.add1file(b_zipped_blob, v_file_name, v_new_blob);
     zz_zip.finish_zip(b_zipped_blob);
     zip_files := zz_zip.get_file_list( b_zipped_blob );
     for i in zip_files.first() .. zip_files.last
     loop
          dbms_output.put_line( zip_files( i ) );
     end loop;
     dbms_output.put_line('--');
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     FileA.txt
     FileB.txt
     FileC.txt
     -- Save zip in t1
     insert into t1 values ('ZipZZ.zip',b_zipped_blob);
     commit;
     -- Create new zip file and add ZipZZ.zip and FileD.txt
     b_zipped_blob := null;
     select file_name, file_blob into v_file_name, v_new_blob
          from t1 where file_name = 'ZipZZ.zip';
     zz_zip.add1file(b_zipped_blob, v_file_name, v_new_blob);
     select file_name, file_blob into v_file_name, v_new_blob
          from t1 where file_name = 'tmp\FileD.txt';
     zz_zip.add1file(b_zipped_blob, v_file_name, v_new_blob);
     zz_zip.finish_zip(b_zipped_blob);
     zip_files := zz_zip.get_file_list( b_zipped_blob );
     for i in zip_files.first() .. zip_files.last
     loop
          dbms_output.put_line( zip_files( i ) );
     end loop;
     dbms_output.put_line('--');
     /* output
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     tmp\FileD.txt
     -- save ZipXX.zip
     insert into t1 values ('ZipXX.zip',b_zipped_blob);
     commit;
end;
--Test with other zip files.
insert into t1 values ('File1.zip',null);
commit;
-- I've created a small zip file File1.zip using Winzip. It contains only 1 small text file File1.txt
-- I use toad to insert into blob column.
declare
v_new_blob blob;
v_file_name varchar2(100);
v_buffer_raw raw(1000);
v_length integer;
b_zipped_blob BLOB;
zip_files zz_zip.file_list;
t_file blob;
begin
     select file_name, file_blob into v_file_name, v_new_blob
          from t1 where file_name = 'File1.zip';
     dbms_output.put_line('Blob length: '||dbms_lob.getlength(v_new_blob));
     zz_zip.add1file(b_zipped_blob, v_file_name, v_new_blob);
     zz_zip.finish_zip(b_zipped_blob);
     zip_files := zz_zip.get_file_list( b_zipped_blob );
     for i in zip_files.first() .. zip_files.last
     loop
          dbms_output.put_line( zip_files( i ) );
     end loop;
     dbms_output.put_line('--');
     -- save new file as Zip1.zip
     insert into t1 values ('ZipFF.zip',b_zipped_blob);
     commit;
end;
     Output
     Blob length: 1855
     File1.zip
     File1.txt
Now, new zip file contains both File1.zip and File.txt.
My expected result was just File1.zip
Thanks.

Your first example looks like I would expect (or do I miss something?).
Your second example is strange, but it I can't reproduce it. If I zip a zipfile with my package it contains only the zipfile:
declare
my_zip blob;
new_zip blob;
zip_files as_zip.file_list;
function file2blob(
    p_dir in varchar2
, p_file_name in varchar2
    return blob
is
    file_lob bfile;
    file_blob blob;
begin
    file_lob := bfilename( p_dir
                         , p_file_name
    dbms_lob.open( file_lob
                 , dbms_lob.file_readonly
    dbms_lob.createtemporary( file_blob
                            , true
    dbms_lob.loadfromfile( file_blob
                         , file_lob
                         , dbms_lob.lobmaxsize
    dbms_lob.close( file_lob );
    return file_blob;
exception
    when others
    then
      if dbms_lob.isopen( file_lob ) = 1
      then
        dbms_lob.close( file_lob );
      end if;
      if dbms_lob.istemporary( file_blob ) = 1
      then
        dbms_lob.freetemporary( file_blob );
      end if;
      raise;
end;
begin
my_zip := file2blob( 'MY_DIR', 'as_fop.zip' );
dbms_output.put_line('zip file to start with');
zip_files := as_zip.get_file_list( my_zip );
for i in zip_files.first() .. zip_files.last
loop
    dbms_output.put_line( zip_files( i ) );
end loop;
dbms_output.put_line('--');
-- now create a new zip file containing the existing zip file as_fop.zip
as_zip.add1file( new_zip, 'as_fop.zip', my_zip );
as_zip.finish_zip( new_zip );
-- see what's in the new zip
dbms_output.put_line('zip file containing a zipfile');
zip_files := as_zip.get_file_list( new_zip );
for i in zip_files.first() .. zip_files.last
loop
    dbms_output.put_line( zip_files( i ) );
end loop;
dbms_output.put_line('--');
end;
Connected to:
Oracle Database 10g Express Edition Release 10.2.0.1.0 - Production
zip file to start with
as_fop.sql
zip file containing a zipfile
as_fop.zip
PL/SQL procedure successfully completed.
ANTON@XE>Anton

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